Question

$\int \frac{1+x+x^2}{x^3(1+x)}+\frac{2x-1}{(x+1{)}^2}+\frac{4-5\sin x}{{\cos }^{2}x}+\frac{1}{{\sin }^2{\cos }^{2}x}+\sin 2x\cos 3x.$ is equal to $.-\frac{1}{x}+klog(x+1)+\frac{n}{(x+1)}+m\tan x+p\sec x+\frac{1}{h}\cos 5x+\frac{1}{2}\cos x-r\cot 2x.$.Find $k+n+m+p+h+r$?

Answer 1

Let $I=\int (\frac{1+x+x^2}{x^3(1+x)}+\frac{2x-1}{{(x+1)}^2}+\frac{4-5\sin x}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x{\cos }^{2}x}+\sin 2x\cos 3x)dx$
$\int \frac{1+x+x^2}{x^3(1+x)}dx=\int (\frac{1}{x^2}+\frac{1}{1+x})dx=\frac{-1}{x}+\log (x+1)$
$\int \frac{2x-1}{{(x+1)}^2}dx=\int \frac{2(x+1)-3}{{(x+1)}^2}dx=\int (\frac{2}{(x+1)}-\frac{3}{{(x+1)}^2})dx=2\log (x+1)+\frac{3}{(x+1)}$
$\int \frac{4-5\sin x}{{\cos }^{2}x}dx=\int (\frac{4}{{\cos }^{2}x}-\frac{5\sin x}{{\cos }^{2}x})dx=\int (4{\sec }^{2}x-5\sec x\tan x)dx=4\tan x-5\tan x$
$\int \sin 2x\cos 3xdx=\frac{1}{2}\int (\sin 5x-\sin x)=\frac{-1}{5}\cos 5x+\frac{1}{\cos x}$
$\int \frac{1}{{\sin }^{2}x{\cos }^{2}x}dx=\int (\frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x})dx=\int ({\sec }^{2}x+{\csc }^{2}x)dx=\tan x-\cot x$
$\therefore k+n+m+P+h+r=17$