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# Find the integral $\int\frac{x^2}{\sqrt{x^2+6}}dx$

## Step-by-step Solution

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###  Videos

$-9\ln\left(\sqrt{x^2+6}+x\right)+6\ln\left(\frac{\sqrt{6}}{6}\sqrt{x^2+6}+\frac{\sqrt{6}}{6}x\right)+\frac{1}{2}\sqrt{x^2+6}x+C_1$
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##  Step-by-step Solution 

Problem to solve:

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

Specify the 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

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

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

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

Find the derivative

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

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

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

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 ${f'(x) = sec^2(x)\cdot D_x(x)}$

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

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

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

$dx=\sqrt{6}\sec\left(\theta \right)^2d\theta$
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 the polynomial $6\tan\left(\theta \right)^2+6$ by it's greatest common factor (GCF): $6$

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

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

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

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

$\int\frac{6\sqrt{6}\tan\left(\theta \right)^2\sec\left(\theta \right)^2}{\sqrt{6}\sqrt{\sec\left(\theta \right)^2}}d\theta$
Why is tan(x)^2+1 = sec(x)^2 ?
7

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

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

Simplify $\sqrt{\sec\left(\theta \right)^2}$ 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 $\frac{1}{2}$

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

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

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

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

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

Divide $1$ by $\sqrt{6}$

$6\sqrt{6}\int\frac{\sqrt{6}}{6}\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$
10

Simplify the expression inside the integral

$6\sqrt{6}\int\frac{\sqrt{6}}{6}\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$
11

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

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

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

$6\int\left(\sec\left(\theta \right)^2-1\right)\sec\left(\theta \right)d\theta$
12

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

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

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

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

$6\int\left(\sec\left(\theta \right)^{3}-\sec\left(\theta \right)\right)d\theta$
13

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

$6\int\left(\sec\left(\theta \right)^{3}-\sec\left(\theta \right)\right)d\theta$
14

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

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

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

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

Multiply the single term $6$ by each term of the polynomial $\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$

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

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

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

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 $x=\theta$

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

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

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

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

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

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

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

Simplify the expression inside the integral

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

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

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

Solve the product $-6\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{3-1}+\frac{3-2}{3-1}\int\sec\left(\theta \right)d\theta\right)$

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

Solve the product $\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)$

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

Simplify the fraction $-6\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}\right)$

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

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

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

Combining like terms $\sqrt{x^2+6}x$ and $-\frac{1}{2}\sqrt{x^2+6}x$

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

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

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

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

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

Simplify the expression inside the integral

$\frac{1}{2}\sqrt{x^2+6}x-3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+6\ln\left(\frac{\sqrt{6}}{6}\sqrt{x^2+6}+\frac{\sqrt{6}}{6}x\right)$
15

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

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

Gather the results of all integrals

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

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

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

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 $x$

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

Simplify the expression inside the integral

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

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

$-6\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$
18

Gather the results of all integrals

$6\ln\left(\frac{\sqrt{6}}{6}\sqrt{x^2+6}+\frac{\sqrt{6}}{6}x\right)-3\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}\sqrt{x^2+6}x-6\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)$
19

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

$6\ln\left(\frac{\sqrt{6}}{6}\sqrt{x^2+6}+\frac{\sqrt{6}}{6}x\right)-9\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}\sqrt{x^2+6}x$
20

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

$6\ln\left(\frac{\sqrt{6}}{6}\sqrt{x^2+6}+\frac{\sqrt{6}}{6}x\right)-9\ln\left(\frac{\sqrt{6}}{6}\left(\sqrt{x^2+6}+x\right)\right)+\frac{1}{2}\sqrt{x^2+6}x+C_0$
21

Simplify the expression by applying logarithm properties

$-9\ln\left(\sqrt{x^2+6}+x\right)+6\ln\left(\frac{\sqrt{6}}{6}\sqrt{x^2+6}+\frac{\sqrt{6}}{6}x\right)+\frac{1}{2}\sqrt{x^2+6}x+C_1$

$-9\ln\left(\sqrt{x^2+6}+x\right)+6\ln\left(\frac{\sqrt{6}}{6}\sqrt{x^2+6}+\frac{\sqrt{6}}{6}x\right)+\frac{1}{2}\sqrt{x^2+6}x+C_1$

##  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

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### Main topic:

Differential Calculus

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