Question

Evaluate ∫1−1x+|x|+1x2+2|x|+1dx

Answer 1

${}_{-1}^1\int \frac{x+\left|x\right|+1}{x^2+2\left|x\right|+1}dx$

$\left|x\right|$ changes sign i.e makes -ve into +ve quantity

so we'll breah the integration in 2 parts.

${}_{-1}^0\frac{x-x+1}{x^2-2x+1}dx{+}_0^1\int \frac{x+x+1}{x^2+2x+1}dx$

${}_{-1}^0\int \frac{1}{(x-1{)}^2}dx{+}_0^1\int \frac{2x+1}{(x+1{)}^2}dx$

${}_{-1}^0\frac{1}{(x-1{)}^2}dx+2_0^1\int \frac{(x+1)}{(x+1{)}^2}dx$

$(x-1)=t$

$dx=dt$

$\int \frac{1}{t^2}dt$

$=-\frac{1}{t}$

$-\frac{1}{(x-1)}{|}_{-1}^0$

$(x+1)=t$

$dx=dt$

$\int \frac{1}{t}dt$

$=lnt$

$ln(x+1){|}_0^1$

$x+1=t$

$dx=dt$

$\int \frac{1}{t^2}dt$

$=-\frac{1}{t}$

$\downarrow$

$\frac{1}{x+1}{|}_0^1$

$\Rightarrow -\frac{1}{(0-1)}+\frac{1}{(-1-1)}+ln2-ln1-[\frac{1}{2}-\frac{1}{1})$

$1-\frac{1}{2}+ln2-\frac{1}{2}+1$

$=1+ln2$