# Step-by-step Solution

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## Final Answer

$-3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}x\sqrt{x^2+6}+C_0$

## Step-by-step Solution

Problem to solve:

$\int\frac{x^2}{\sqrt{x^2+6}}dx$

Solving method

1

We can solve the integral $\int\frac{x^2}{\sqrt{x^2+6}}dx$ by applying integration method of trigonometric substitution using the substitution

$x=\frac{6}{\sqrt{6}}\tan\left(\theta \right)$

Differentiate both sides of the equation $x=\frac{6}{\sqrt{6}}\tan\left(\theta \right)$

$dx=\frac{d}{d\theta}\left(\frac{6}{\sqrt{6}}\tan\left(\theta \right)\right)$

Find the derivative

$\frac{d}{d\theta}\left(\frac{6}{\sqrt{6}}\tan\left(\theta \right)\right)$

The derivative of a function multiplied by a constant ($\frac{6}{\sqrt{6}}$) is equal to the constant times the derivative of the function

$\frac{6}{\sqrt{6}}\frac{d}{d\theta}\left(\tan\left(\theta \right)\right)$

The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if ${f(x) = tan(x)}$, then ${f'(x) = sec^2(x)\cdot D_x(x)}$

$\frac{6}{\sqrt{6}}\sec\left(\theta \right)^2\frac{d}{d\theta}\left(\theta \right)$

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

$\frac{6}{\sqrt{6}}\sec\left(\theta \right)^2$
2

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

$dx=\frac{6}{\sqrt{6}}\sec\left(\theta \right)^2d\theta$
3

Substituting in the original integral, we get

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\sqrt{6\tan\left(\theta \right)^2+6}}d\theta$
4

Factor by the greatest common divisor $6$

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\sqrt{6\left(\tan\left(\theta \right)^2+1\right)}}d\theta$

$\int\frac{6}{\sqrt{6}}\left(\frac{\frac{6}{\sqrt{6}}^2\tan\left(\theta \right)^2}{\sqrt{\left(\frac{6}{\sqrt{6}}\tan\left(\theta \right)\right)^2+6}}\right)\sec\left(\theta \right)^2d\theta$

Calculate the power $\frac{6}{\sqrt{6}}^2$

$\int\frac{6}{\sqrt{6}}\left(\frac{6\tan\left(\theta \right)^2}{\sqrt{\left(\frac{6}{\sqrt{6}}\tan\left(\theta \right)\right)^2+6}}\right)\sec\left(\theta \right)^2d\theta$

Multiplying the fraction by $\frac{6}{\sqrt{6}}$

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\sqrt{\left(\frac{6}{\sqrt{6}}\tan\left(\theta \right)\right)^2+6}}d\theta$

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\sqrt{\frac{6}{\sqrt{6}}^2\tan\left(\theta \right)^2+6}}d\theta$

Calculate the power $\frac{6}{\sqrt{6}}^2$

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\sqrt{6\tan\left(\theta \right)^2+6}}d\theta$

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\sqrt{6}\sqrt{\tan\left(\theta \right)^2+1}}d\theta$

Calculate the power $\sqrt{6}$

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\frac{6}{\sqrt{6}}\sqrt{\tan\left(\theta \right)^2+1}}d\theta$
5

The power of a product is equal to the product of it's factors raised to the same power

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\frac{6}{\sqrt{6}}\sqrt{\tan\left(\theta \right)^2+1}}d\theta$
6

Applying the trigonometric identity: $\tan(x)^2+1=\sec(x)^2$

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\frac{6}{\sqrt{6}}\sec\left(\theta \right)}d\theta$
7

Taking the constant ($6\sqrt{6}$) out of the integral

$6\sqrt{6}\int\frac{\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\frac{6}{\sqrt{6}}\sec\left(\theta \right)}d\theta$
8

Simplify the fraction by $\sec\left(\theta \right)$

$6\sqrt{6}\int\frac{\tan\left(\theta \right)^2\sec\left(\theta \right)}{\frac{6}{\sqrt{6}}}d\theta$
9

Rewrite the fraction $\frac{\tan\left(\theta \right)^2\sec\left(\theta \right)}{\frac{6}{\sqrt{6}}}$

$6\sqrt{6}\int\frac{\sqrt{6}}{6}\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$

$6\sqrt{6}\cdot \frac{\sqrt{6}}{6}\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$

Multiply $6\sqrt{6}$ times $\frac{\sqrt{6}}{6}$

$6\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$
10

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$6\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$

Applying the trigonometric identity: $\tan^2(\theta)=\sec(\theta)^2-1$

$6\int\left(\sec\left(\theta \right)^2-1\right)\sec\left(\theta \right)d\theta$

Multiplying polynomials $\sec\left(\theta \right)$ and $\sec\left(\theta \right)^2-1$

$6\int\left(\sec\left(\theta \right)\sec\left(\theta \right)^2-\sec\left(\theta \right)\right)d\theta$

When multiplying exponents with same base you can add the exponents: $\sec\left(\theta \right)\sec\left(\theta \right)^2$

$6\int\left(\sec\left(\theta \right)^{3}-\sec\left(\theta \right)\right)d\theta$

The integral of the sum of two or more functions is equal to the sum of their integrals

$6\left(\int\sec\left(\theta \right)^{3}d\theta+\int-\sec\left(\theta \right)d\theta\right)$

$6\left(\int\sec\left(\theta \right)^3d\theta-\int\sec\left(\theta \right)d\theta\right)$

Solve the product $6\left(\int\sec\left(\theta \right)^3d\theta-\int\sec\left(\theta \right)d\theta\right)$

$6\int\sec\left(\theta \right)^3d\theta-6\int\sec\left(\theta \right)d\theta$
11

Apply the formula: $\int\sec\left(x\right)\tan\left(x\right)^2dx$$=\int\sec\left(x\right)^3dx-\int\sec\left(x\right)dx, where x=\theta 6\int\sec\left(\theta \right)^3d\theta-6\int\sec\left(\theta \right)d\theta Rewrite \sec\left(\theta \right)^3 as the product of two secants 6\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta We can solve the integral \int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta by applying integration by parts method to calculate the integral of the product of two functions, using the following formula \displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du First, identify u and calculate du \begin{matrix}\displaystyle{u=\sec\left(\theta \right)}\\ \displaystyle{du=\sec\left(\theta \right)\tan\left(\theta \right)d\theta}\end{matrix} Now, identify dv and calculate v \begin{matrix}\displaystyle{dv=\sec\left(\theta \right)^2d\theta}\\ \displaystyle{\int dv=\int \sec\left(\theta \right)^2d\theta}\end{matrix} Solve the integral v=\int\sec\left(\theta \right)^2d\theta The integral of \sec(x)^2 is \tan(x) \tan\left(\theta \right) Now replace the values of u, du and v in the last formula 6\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right) Solve the product 6\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right) 6\tan\left(\theta \right)\sec\left(\theta \right)-6\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta Apply the formula: \int\sec\left(x\right)\tan\left(x\right)^2dx$$=\int\sec\left(x\right)^3dx-\int\sec\left(x\right)dx$, where $x=\theta$

$6\tan\left(\theta \right)\sec\left(\theta \right)-6\int\sec\left(\theta \right)^3d\theta+6\int\sec\left(\theta \right)d\theta$

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

$6\tan\left(\theta \right)\sec\left(\theta \right)-6\int\sec\left(\theta \right)^3d\theta+6\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

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

$6\tan\left(\theta \right)\sec\left(\theta \right)-6\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta\right)+6\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

Solve the product $-6\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta\right)$

$6\tan\left(\theta \right)\sec\left(\theta \right)-3\sin\left(\theta \right)\sec\left(\theta \right)^{2}-3\int\sec\left(\theta \right)d\theta+6\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

Simplifying

$3\tan\left(\theta \right)\sec\left(\theta \right)-3\int\sec\left(\theta \right)d\theta+6\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

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

$3\tan\left(\theta \right)\sec\left(\theta \right)-3\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)+6\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

Combining like terms $-3\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$ and $6\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

$3\tan\left(\theta \right)\sec\left(\theta \right)+3\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

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

$3\left(\frac{x}{\frac{6}{\sqrt{6}}}\right)\left(\frac{\sqrt{x^2+6}}{\frac{6}{\sqrt{6}}}\right)+3\ln\left(\frac{\sqrt{x^2+6}}{\frac{6}{\sqrt{6}}}+\frac{x}{\frac{6}{\sqrt{6}}}\right)$

Multiplying the fraction by $3$

$\frac{3x}{\frac{6}{\sqrt{6}}}\frac{\sqrt{x^2+6}}{\frac{6}{\sqrt{6}}}+3\ln\left(\frac{\sqrt{x^2+6}}{\frac{6}{\sqrt{6}}}+\frac{x}{\frac{6}{\sqrt{6}}}\right)$

Take $\frac{3}{\frac{6}{\sqrt{6}}}$ out of the fraction

$\frac{1}{2}x\sqrt{x^2+6}+3\ln\left(\frac{\sqrt{x^2+6}}{\frac{6}{\sqrt{6}}}+\frac{x}{\frac{6}{\sqrt{6}}}\right)$

Simplifying

$\frac{1}{2}x\sqrt{x^2+6}+3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$
12

The integral $6\int\sec\left(\theta \right)^3d\theta$ results in: $\frac{1}{2}x\sqrt{x^2+6}+3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$

$\frac{1}{2}x\sqrt{x^2+6}+3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$
13

Gather the results of all integrals

$3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}x\sqrt{x^2+6}-6\int\sec\left(\theta \right)d\theta$

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

$-6\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

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

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

Simplifying

$-6\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$
14

The integral $-6\int\sec\left(\theta \right)d\theta$ results in: $-6\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$

$-6\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$
15

Gather the results of all integrals

$3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}x\sqrt{x^2+6}-6\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$
16

Combining like terms $3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$ and $-6\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$

$-3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}x\sqrt{x^2+6}$
17

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

$-3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}x\sqrt{x^2+6}+C_0$

## Final Answer

$-3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}x\sqrt{x^2+6}+C_0$
$\int\frac{x^2}{\sqrt{x^2+6}}dx$

### Main topic:

Integrals of Rational Functions

~ 0.23 s