Question

Integrate the following with respect to x:\

${\displaystyle \frac{\log x}{{(1+\log )}^2}}$

Answer 1

I = ${\displaystyle \int \frac{\log x}{{(1+\log x)}^2} \text{d}x}$

Put t = ${\displaystyle \log x}$

⇒ x = e^t

dx = e^t dt

I = ${\displaystyle \int \frac{\text{t}}{{(1+\text{t})}^2}{\text{e}}^{\text{t}}\cdot \text{dt}}$

I = ${\displaystyle \int {\text{e}}^{\text{t}}\frac{1+\text{t}-1}{{(1+\text{t})}^2} \text{dt}}$

I = ${\displaystyle \int {\text{e}}^{\text{t}}[\frac{1+\text{t}}{{(1+\text{t})}^2}-\frac{1}{{(1+\text{t})}^2}] \text{dt}}$

I = ${\displaystyle \int {\text{e}}^{\text{t}}[\frac{1}{1+\text{t}}-\frac{1}{{(1+\text{t})}^2}] \text{dt}}$

f(t) = ${\displaystyle \frac{1}{1+\text{t}}}$

f'(t) = ${\displaystyle -\frac{1}{{(1+\text{t})}^2}}$

I = ${\displaystyle {\text{e}}^{\text{t}}\cdot \frac{1}{1+\text{t}}+\text{c}}$

= ${\displaystyle x\cdot \frac{1}{1+\log x}+\text{c}}$

I = ${\displaystyle \frac{x}{1+\log x}+\text{c}}$