Question

Solve:${\int }_1^2\frac{1}{x(1+x^2)}dx$

Answer 1

Given, ${\int }_1^2\frac{1}{x(1+x^2)}dx$

We can write $\frac{1}{x(1+x^2)}$ as $\frac{1}{x}-\frac{x}{1+x^2}$

$\therefore {\int }_1^2\frac{1}{x(1+x^2)}dx={\int }_1^2(\frac{1}{x}-\frac{x}{1+x^2})dx$

Now,

${\int }_1^2(\frac{1}{x}-\frac{x}{1+x^2})dx={\int }_1^2\frac{1}{x}dx-{\int }_1^2\frac{x}{1+x^2}dx$

So, ${\int }_1^2\frac{1}{x}dx={\left[\ln \left|x\right|\right]}_1^2+C_1$

for, ${\int }_1^2\frac{x}{1+x^2}dx$,

Let, $u=1+x^2\Rightarrow \frac{du}{dx}=2x\Rightarrow du=2xdx\Rightarrow \frac{1}{2}du=xdx$

Substituting these values we get,

${\int }_1^2\frac{x}{1+x^2}dx={\int }_1^2\frac{1}{2u}du=\frac{1}{2}{\int }_1^2\frac{1}{u}du=\frac{1}{2}{\left[\ln \left|u\right|\right]}_1^2+C_2$

$\therefore {\int }_1^2\frac{1}{x}dx-{\int }_1^2\frac{x}{1+x^2}dx={\left[\ln \left|x\right|\right]}_1^2-\frac{1}{2}{\left[\ln \left|u\right|\right]}_1^2+C={\left[\ln \left|x\right|\right]}_1^2-\frac{1}{2}{\left[\ln |1+x^2|\right]}_1^2+C=[\ln 2-\ln 1]-\frac{1}{2}\left[\ln (1+2^2)-\ln (1+1^2)\right]+C=[\ln 2-\ln 1]-\frac{1}{2}[\ln 5-\ln 2]+C=\ln \left(\frac{2}{1}\right)-\frac{1}{2}\ln \left(\frac{5}{2}\right)+C\therefore {\int }_1^2\frac{1}{x(1+x^2)}dx=\ln \left(\frac{2}{1}\right)-\frac{1}{2}\ln \left(\frac{5}{2}\right)+C.$