Question

Need some help with indefinite integrals pleasee

Answer 1

Observe that $(x^2+2x+2)'=2x+2$.

You have that $\int \frac{2x}{x^2+2x+2}\; dx=\int \frac{2x+2-2}{x^2+2x+2}\; dx=\int \frac{2x+2}{x^2+2x+2}\; dx+\int \frac{1}{x^2+2x+2}\; dx=I_1+2I_2$.

To compute $I_1$ you set $y=x^2+2x+2$, so $dy=2x+2\; dx$. Therefore, $I_1=\int \frac{1}{y}\; dy=\ln |y|=\ln |x^2+2x+2|=\ln (x^2+2x+2)$.

For $I_2$ observe that $x^2+2x+2=(x+1)^2+1$. Let $y=x+1$. Then $dy=dx$ and $I_2=\int \frac{1}{y^2+1}=\arctan y=\arctan (x+1)$.

So, $\int \frac{2x}{x^2+2x+2}=\ln (x^2+2x+2)+2\arctan (x+1)+c$.