Exercise

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

Step-by-step Solution

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

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

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

Simplifying

$\int6\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$
5

The integral of a function times a constant ($6$) is equal to the constant times the integral of the function

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

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

$6\int\left(\sec\left(\theta \right)^2-1\right)\sec\left(\theta \right)d\theta$
7

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

$6\int\left(\sec\left(\theta \right)^{3}-\sec\left(\theta \right)\right)d\theta$
8

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

$6\int\sec\left(\theta \right)^{3}d\theta-6\int\sec\left(\theta \right)d\theta$
9

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

$\frac{1}{2}x\sqrt{x^2+6}+3\ln\left(\frac{\sqrt{x^2+6}+x}{\sqrt{6}}\right)$
10

Gather the results of all integrals

$3\ln\left|\frac{\sqrt{x^2+6}+x}{\sqrt{6}}\right|+\frac{1}{2}x\sqrt{x^2+6}-6\int\sec\left(\theta \right)d\theta$
11

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

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

Gather the results of all integrals

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

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

$-3\ln\left|\frac{\sqrt{x^2+6}+x}{\sqrt{6}}\right|+\frac{1}{2}x\sqrt{x^2+6}$
14

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

$-3\ln\left|\frac{\sqrt{x^2+6}+x}{\sqrt{6}}\right|+\frac{1}{2}x\sqrt{x^2+6}+C_0$
15

Simplify the expression by applying the property of the logarithm of a quotient

$-3\ln\left|\sqrt{x^2+6}+x\right|+\frac{1}{2}x\sqrt{x^2+6}+C_1$

Final answer to the exercise

$-3\ln\left|\sqrt{x^2+6}+x\right|+\frac{1}{2}x\sqrt{x^2+6}+C_1$

Try other ways to solve this exercise

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