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Solve the integral of logarithmic functions $\int\ln\left(\sqrt{x}+\sqrt{1+x}\right)dx$

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Final Answer

$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}+C_0$
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Step-by-step Solution

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1

We can solve the integral $\int\ln\left(\sqrt{x}+\sqrt{1+x}\right)dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{\sqrt{x}+\sqrt{1+x}}\frac{d}{dx}\left(\sqrt{x}+\sqrt{1+x}\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{1}{\sqrt{x}+\sqrt{1+x}}\left(\frac{d}{dx}\left(\sqrt{x}\right)+\frac{d}{dx}\left(\sqrt{1+x}\right)\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{1}{\sqrt{x}+\sqrt{1+x}}\left(\frac{d}{dx}\left(\sqrt{x}\right)+\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}\frac{d}{dx}\left(1+x\right)\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{1}{\sqrt{x}+\sqrt{1+x}}\left(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}\frac{d}{dx}\left(1+x\right)\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{1}{\sqrt{x}+\sqrt{1+x}}\left(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}\left(\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)\right)\right)$

The derivative of the constant function ($1$) is equal to zero

$\frac{1}{\sqrt{x}+\sqrt{1+x}}\left(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}\frac{d}{dx}\left(x\right)\right)$

The derivative of the linear function is equal to $1$

$\frac{1}{\sqrt{x}+\sqrt{1+x}}\left(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}\right)$

Multiplying the fraction by $\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}$

$\frac{\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}}{\sqrt{x}+\sqrt{1+x}}$

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{1+x}}}{\sqrt{x}+\sqrt{1+x}}$

The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors

$L.C.M.=2\sqrt{x}\sqrt{1+x}$

Obtained the least common multiple (LCM), we place it as the denominator of each fraction, and in the numerator of each fraction we add the factors that we need to complete

$\frac{\sqrt{1+x}}{2\sqrt{x}\sqrt{1+x}}+\frac{\sqrt{x}}{2\sqrt{x}\sqrt{1+x}}$

Combine and simplify all terms in the same fraction with common denominator $2\sqrt{x}\sqrt{1+x}$

$\frac{\frac{\sqrt{1+x}+\sqrt{x}}{2\sqrt{x}\sqrt{1+x}}}{\sqrt{x}+\sqrt{1+x}}$

Divide fractions $\frac{\frac{\sqrt{1+x}+\sqrt{x}}{2\sqrt{x}\sqrt{1+x}}}{\sqrt{x}+\sqrt{1+x}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$\frac{\sqrt{1+x}+\sqrt{x}}{2\sqrt{x}\sqrt{1+x}\left(\sqrt{x}+\sqrt{1+x}\right)}$

Simplify the fraction $\frac{\sqrt{1+x}+\sqrt{x}}{2\sqrt{x}\sqrt{1+x}\left(\sqrt{x}+\sqrt{1+x}\right)}$ by $\sqrt{1+x}+\sqrt{x}$

$\frac{1}{2\sqrt{x}\sqrt{1+x}}$
2

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=\ln\left(\sqrt{x}+\sqrt{1+x}\right)}\\ \displaystyle{du=\frac{1}{2\sqrt{x}\sqrt{1+x}}dx}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=1dx}\\ \displaystyle{\int dv=\int 1dx}\end{matrix}$
4

Solve the integral

$v=\int1dx$
5

The integral of a constant is equal to the constant times the integral's variable

$x$

Simplify the fraction by $x$

$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)-\int\frac{\sqrt{x}}{2\sqrt{1+x}}dx$
6

Now replace the values of $u$, $du$ and $v$ in the last formula

$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)-\int\frac{\sqrt{x}}{2\sqrt{1+x}}dx$

Take the constant $\frac{1}{2}$ out of the integral

$- \frac{1}{2}\int\frac{\sqrt{x}}{\sqrt{1+x}}dx$

Multiply $-1$ times $\frac{1}{2}$

$-\frac{1}{2}\int\frac{\sqrt{x}}{\sqrt{1+x}}dx$

We can solve the integral $\int\frac{\sqrt{x}}{\sqrt{1+x}}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $\sqrt{1+x}$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=\sqrt{1+x}$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}dx$

Isolate $dx$ in the previous equation

$\frac{du}{\frac{1}{2}\left(1+x\right)^{-\frac{1}{2}}}=dx$

Rewriting $x$ in terms of $u$

$x=u^{2}-1$

Substituting $u$, $dx$ and $x$ in the integral and simplify

$-\frac{1}{2}\int2\sqrt{u^{2}-1}du$

The integral of a function times a constant ($2$) is equal to the constant times the integral of the function

$-\int\sqrt{u^{2}-1}du$

We can solve the integral $-\int\sqrt{u^{2}-1}du$ by applying integration method of trigonometric substitution using the substitution

$u=\sec\left(\theta \right)$

Now, in order to rewrite $d\theta$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=\sec\left(\theta \right)\tan\left(\theta \right)d\theta$

Substituting in the original integral, we get

$-\int\sqrt{\sec\left(\theta \right)^{2}-1}\sec\left(\theta \right)\tan\left(\theta \right)d\theta$

Apply the trigonometric identity: $\sec\left(\theta \right)^2-1$$=\tan\left(\theta \right)^2$, where $x=\theta $

$-\int\sqrt{\tan\left(\theta \right)^2}\sec\left(\theta \right)\tan\left(\theta \right)d\theta$

Simplify $\sqrt{\tan\left(\theta \right)^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$

$-\int\tan\left(\theta \right)\sec\left(\theta \right)\tan\left(\theta \right)d\theta$

When multiplying two powers that have the same base ($\tan\left(\theta \right)$), you can add the exponents

$-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$

We identify that the integral has the form $\int\tan^m(x)\sec^n(x)dx$. If $n$ is odd and $m$ is even, then we need to express everything in terms of secant, expand and integrate each function separately

$-\int\left(\sec\left(\theta \right)^2-1\right)\sec\left(\theta \right)d\theta$

Multiply the single term $\sec\left(\theta \right)$ by each term of the polynomial $\left(\sec\left(\theta \right)^2-1\right)$

$-\int\left(\sec\left(\theta \right)^{3}-\sec\left(\theta \right)\right)d\theta$

Expand the integral $\int\left(\sec\left(\theta \right)^{3}-\sec\left(\theta \right)\right)d\theta$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$-\int\sec\left(\theta \right)^{3}d\theta-\int-\sec\left(\theta \right)d\theta$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$-\int\sec\left(\theta \right)^{3}d\theta+1\int\sec\left(\theta \right)d\theta$

Any expression multiplied by $1$ is equal to itself

$-\int\sec\left(\theta \right)^{3}d\theta+\int\sec\left(\theta \right)d\theta$

The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$

$-\int\sec\left(\theta \right)^{3}d\theta+\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

Express the variable $\theta$ in terms of the original variable $x$

$-\int\sec\left(\theta \right)^{3}d\theta+\ln\left(u+\sqrt{u^{2}-1}\right)$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1+x}$

$-\int\sec\left(\theta \right)^{3}d\theta+\ln\left(\sqrt{1+x}+\sqrt{1+x-1}\right)$

Subtract the values $1$ and $-1$

$-\int\sec\left(\theta \right)^{3}d\theta+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Rewrite $\sec\left(\theta \right)^{3}$ as the product of two secants

$-\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

We can solve the integral $\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=\sec\left(\theta \right)}\\ \displaystyle{du=\sec\left(\theta \right)\tan\left(\theta \right)d\theta}\end{matrix}$

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sec\left(\theta \right)^2d\theta}\\ \displaystyle{\int dv=\int \sec\left(\theta \right)^2d\theta}\end{matrix}$

Solve the integral

$v=\int\sec\left(\theta \right)^2d\theta$

The integral of $\sec(x)^2$ is $\tan(x)$

$\tan\left(\theta \right)$

Now replace the values of $u$, $du$ and $v$ in the last formula

$-\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Multiply the single term $-1$ by each term of the polynomial $\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$

$-\tan\left(\theta \right)\sec\left(\theta \right)+\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Express the variable $\theta$ in terms of the original variable $x$

$-\sqrt{u^{2}-1}u+\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1+x}$

$-\sqrt{1+x-1}\sqrt{1+x}+\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Subtract the values $1$ and $-1$

$-\sqrt{x}\sqrt{1+x}+\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Apply the formula: $\int\sec\left(\theta \right)\tan\left(\theta \right)^2dx$$=\int\sec\left(\theta \right)^3dx-\int\sec\left(\theta \right)dx$, where $x=\theta $

$-\sqrt{x}\sqrt{1+x}+\int\sec\left(\theta \right)^3d\theta-\int\sec\left(\theta \right)d\theta+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$

$-\sqrt{x}\sqrt{1+x}+\int\sec\left(\theta \right)^3d\theta-\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Express the variable $\theta$ in terms of the original variable $x$

$-\sqrt{x}\sqrt{1+x}+\int\sec\left(\theta \right)^3d\theta-\ln\left(u+\sqrt{u^{2}-1}\right)+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1+x}$

$-\sqrt{x}\sqrt{1+x}+\int\sec\left(\theta \right)^3d\theta-\ln\left(\sqrt{1+x}+\sqrt{1+x-1}\right)+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Subtract the values $1$ and $-1$

$-\sqrt{x}\sqrt{1+x}+\int\sec\left(\theta \right)^3d\theta-\ln\left(\sqrt{1+x}+\sqrt{x}\right)+\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

Cancel like terms $-\ln\left(\sqrt{1+x}+\sqrt{x}\right)$ and $\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

$-\sqrt{x}\sqrt{1+x}+\int\sec\left(\theta \right)^3d\theta$

Simplify the integral $\int\sec\left(\theta \right)^3d\theta$ applying the reduction formula, $\displaystyle\int\sec(x)^{n}dx=\frac{\sin(x)\sec(x)^{n-1}}{n-1}+\frac{n-2}{n-1}\int\sec(x)^{n-2}dx$

$-\sqrt{x}\sqrt{1+x}+\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{3-1}+\frac{3-2}{3-1}\int\sec\left(\theta \right)d\theta$

Simplify the expression inside the integral

$-\sqrt{x}\sqrt{1+x}+\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta$

Express the variable $\theta$ in terms of the original variable $x$

$-\sqrt{x}\sqrt{1+x}+\frac{\sqrt{u^{2}-1}u}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1+x}$

$-\sqrt{x}\sqrt{1+x}+\frac{\sqrt{1+x-1}\sqrt{1+x}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta$

Subtract the values $1$ and $-1$

$-\sqrt{x}\sqrt{1+x}+\frac{\sqrt{x}\sqrt{1+x}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta$

Simplify the expression inside the integral

$-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\int\sec\left(\theta \right)d\theta$

The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$

$-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

Express the variable $\theta$ in terms of the original variable $x$

$-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(u+\sqrt{u^{2}-1}\right)$

Replace $u$ with the value that we assigned to it in the beginning: $\sqrt{1+x}$

$-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{1+x-1}\right)$

Subtract the values $1$ and $-1$

$-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)$
7

The integral $-\int\frac{\sqrt{x}}{2\sqrt{1+x}}dx$ results in: $-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)$

$-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)$
8

Gather the results of all integrals

$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}$
9

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}+C_0$

Final Answer

$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}+C_0$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of ln(x^0.5+(1+x)^0.5)dx using basic integralsSolve integral of ln(x^0.5+(1+x)^0.5)dx using u-substitutionSolve integral of ln(x^0.5+(1+x)^0.5)dx using integration by partsSolve integral of ln(x^0.5+(1+x)^0.5)dx using tabular integration

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Function Plot

Plotting: $x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}+C_0$

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log
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|◻|
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>=
<=
sin
cos
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coth
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asinh
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acoth
asech
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