Question

Evaluate this integral x(x²+4)dx / x⁴+9

Answer 1

$$\int \dfrac{x(x^2+4)}{x^4+9}dx$$

Substitute $x^2=u, 2xdx=du$

$$\int \dfrac{x(x^2+4)}{x^4+9}dx=\dfrac{1}{2}\int \dfrac{u+4}{u^2+9}du$$

$$=\dfrac{1}{2}\int \dfrac{u}{u^2+9}du+2\int \dfrac{du}{u^2+9}$$

$$\dfrac{1}{2}\int \dfrac{u}{u^2+9}du$$

$$t=u^2+9, dt=2udu$$

$$\dfrac{1}{2}\int \dfrac{u}{u^2+9}du=\dfrac{1}{4}\int \dfrac{dt}{t}=$$

$$=\dfrac{1}{4}\ln |t|+C_1=\dfrac{1}{4}\ln(u^2+9)+C_1$$

$$=\dfrac{1}{4}\ln(x^4+9)+C_1$$

$$2\int \dfrac{du}{u^2+9}$$

$$s=\dfrac{u}{3}, ds=\dfrac{1}{3}du$$

$$2\int \dfrac{du}{u^2+9}=2\int \dfrac{3ds}{9s^2+9}=\dfrac{2}{3}\int \dfrac{ds}{1+s^2}$$

$$=\dfrac{2}{3}\tan^{-1}(s)+C_2=\dfrac{2}{3}\tan^{-1}(\dfrac{u}{3})+C_2$$

$$=\dfrac{2}{3}\tan^{-1}(\dfrac{x^2}{3})+C_2$$

Then

$$\int \dfrac{x(x^2+4)}{x^4+9}dx=\dfrac{1}{4}\ln(x^4+9)+\dfrac{2}{3}\tan^{-1}(\dfrac{x^2}{3})+C$$