Solving an indefinite integral using integration by parts

Question

As I was preparing for a test I came across an integral within the section of integration by parts: $$\int{\frac{dx}{(x^2+4)^3}}$$ Then I tried using an online integral calculator, I got the right solution, but the problem was that for the first part of solving the integral it used reduction formula, which we weren't allowed to use in this example. How can I solve this integral by using only integration by parts or integration by parts in combination with u-substitution?

Answer 1

Hint

$$I=\int{\frac{dx}{(x^2+4)^3}}$$ $$x=2\tan(t) \implies dx=2 \sec ^2(t)\implies I=\frac 1{32}\int \cos^4(t)\,dt$$

Using twice the double angle formulae, $$\cos^4(t)=\frac{3}{8}+\frac{1}{2} \cos (2 t)+\frac{1}{8} \cos (4 t)$$

Answer 2

As you are looking for IBP combined with U-substitution, you can establish a reduction formula;

$$I_n=\int \frac{1}{(x^2+a^2)^n}\,dx$$

$$I_n=\int \frac{x^2+a^2-x^2}{a^2(x^2+a^2)^n}\,dx$$

$$I_n=\frac{1}{a^2}I_{n-1}-\frac{1}{a^2} \int \frac{x^2}{a^2(x^2+a^2)^n}\,dx$$

$$I_n=\frac{1}{a^2}I_{n-1}-\frac{1}{2a^2} \int \frac{2(x)(x)}{a^2(x^2+a^2)^n}\,dx$$

$$I_n=\frac{1}{a^2}I_{n-1}-\frac{1}{2a^2}\left(\frac{-x}{(n-1)(x^2+a^2)^{n-1}}+\int \frac{1}{(n-1)(x^2+a^2)^{n-1}}\,dx\right)$$

$$\boxed{I_n=\frac{1}{a^2}\left(\frac{2n-3}{2n-2}\right)I_{n-1}+\frac{1}{2a^2}\left(\frac{x}{(n-1)(x^2+a^2)^{n-1}}\right)}$$

Since the original question is $I_3$ at $a=2$,

$$\boxed{I_3=\frac{1}{4}\left(\frac{3}{4}\right)I_{2}+\frac{1}{8}\left(\frac{x}{(2)(x^2+a^2)^{2}}\right)}$$

Putting in $I_2$,

$$\boxed{I_2=\frac{1}{4}\left(\frac{1}{2}\right)I_{1}+\frac{1}{8}\left(\frac{x}{(x^2+a^2)}\right)}$$

Putting in $I_1$,

$I_1$ is a well known integral; $$\boxed{I_1=\frac{1}{2} arctan\left(\frac{x}{2} \right)}$$

To conclude;

$$\boxed{I_3=\left(\frac{3}{16}\right)\left[\frac{1}{16} arctan\left(\frac{x}{2} \right)+\frac{1}{64}\frac{x}{(x^2+a^2)}+\frac{x}{(16)(x^2+a^2)^2}\right]+C}$$