Integrate $\int\frac{dx}{(x^2+16)^3}$

Question

Solve the following integral:

$$\int\frac{dx}{(x^2+16)^3}$$

I have no idea what to do here. I think there will be trigonometric substitution but I can't even seem to get started with this problem.

Thanks in advance.

Answer 1

Remember the trick, whenever it is of form $(x^2 + a^2)$ , we generally prefer to substitute $x = a\tan \theta$ or $x = a\cot \theta$ depending upon the conditions.

Here, $\bf{a = 4 }$ and thus, it will be fair enough to substitute, $x = 4\tan \theta$.

$$x = 4\tan \theta \implies dx = 4\sec^2 \theta d\theta$$ The rest has been explained well by other users. Simply find the integral value :

$$\begin{align} &\int\frac{4\sec^2 \theta d\theta}{\left(16 \tan^2 \theta + 16\right)^3}\\ =&\int\frac{4\sec^2 \theta d\theta}{\left(16\sec^2 \theta\right)^3} \end{align}$$

Answer 2

Hint: It seems to be trigonometric substitution. Let $x=4 \tan \theta$. Then the integral becomes $$\int \frac{4 \sec^2 \theta \, d\theta}{(16 \tan^2 \theta + 16)^3}$$

Answer 3

Substitute $x=4\tan\theta$ to get $$\int\frac{4\sec^2 \theta d\theta}{\left(16\sec^2 \theta\right)^3}$$

Answer 4

Hint: Evaluate $I(a)=\displaystyle\int\frac{dx}{x^2+a}~,~$ and then differentiate twice with regard to a.