Answer to Question #131179 in Calculus for mae

Question

1. ∫ lnx/(1+x^2) dx
2. ∫ sinx/cosx(1+cos^2x) dx

Answer 1

1. $\int \ln{\frac{x}{x^2+1}} = |d(\ln{\frac{x}{x^2+1}})= \frac{x^2+1}{x} \frac{(x^2+1)-x\cdot 2x}{(x^2+1)^2}dx= \frac{1-x^2}{(1+x^2)x}dx| = ln(\frac{x}{x^2+1}) \cdot x - \int \frac{1-x^2}{1+x^2}dx = ln(\frac{x}{x^2+1}) \cdot x - \int \frac{1}{1+x^2}dx + \int \frac{x^2}{1+x^2}dx =ln(\frac{x}{x^2+1}) \cdot x -\arctan{x} + \int \frac{x^2+1-1}{1+x^2}dx =ln(\frac{x}{x^2+1}) \cdot x -\arctan{x} + \int \frac{x^2+1}{1+x^2}dx - \int \frac{1}{x^2+1}dx = ln(\frac{x}{x^2+1}) \cdot x -2 \arctan{x}+ x +C$

2. $\int \frac{sinx}{cosx(1+cos^2x)}dx=\int sinx (\frac{1}{cosx}-\frac{cosx}{1+cos^2x})dx=$

$=\int \frac{sinx}{cosx}dx-\int \frac{sinx\cdot cosx}{1+cos^2x}dx=$

$=\int -\frac{1}{cosx}dcosx+\int \frac{cosx}{1+cos^2x}dcosx=$

$=-ln(cosx)+\int \frac{1}{2(1+cos^2x)}dcos^2x=$

$=-ln(cosx)+ln(1+cos^2x) +C$