Find the integral int(((x^2)/((x^2+6)^0.5)))dx

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

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

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 $

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

12

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

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

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

Subtract the values $3$ and $-2$

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

Any expression to the power of $1$ is equal to that same expression

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

13

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

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

14

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

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

Taking the derivative of secant function: $\frac{d}{dx}\left(\sec(x)\right)=\sec(x)\cdot\tan(x)\cdot D_x(x)$

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

15

First, identify $u$ and calculate $du$

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

16

Now, identify $dv$ and calculate $v$

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

17

Solve the integral

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

18

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

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

19

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

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

20

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

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