# Integration

• Aug 1st 2013, 06:32 AM
srirahulan
Integration
$\int[\frac{ln(x+1)}{x+1}]$ How can i solve this for simple way. i think a solution for a longtime but i cannot get anything.please give me solution?????
• Aug 1st 2013, 07:03 AM
JJacquelin
Re: Integration
You forget dx in the integral (very important. You need it to the change of variable)
Let y=ln(x+1)
dy=dx/(x+1)
• Aug 1st 2013, 07:28 AM
Plato
Re: Integration
Quote:

Originally Posted by srirahulan
$\int[\frac{ln(x+1)}{x+1}]$ How can i solve this for simple way. i think a solution for a longtime but i cannot get anything.please give me solution?????

What is the derivative of $\tfrac{1}{2}\ln^2(x+1)~?$
• Aug 1st 2013, 08:17 PM
ibdutt
Re: Integration
• Aug 1st 2013, 08:59 PM
Prove It
Re: Integration
Quote:

Originally Posted by srirahulan
$\int[\frac{ln(x+1)}{x+1}]$ How can i solve this for simple way. i think a solution for a longtime but i cannot get anything.please give me solution?????

No, you won't be "given a solution", but you can be given some hints. First of all, having "x + 1" in both the numerator and denominator makes things look convoluted. So a substitution \displaystyle \begin{align*} u = x + 1 \implies du = dx \end{align*} would clean things up a bit, giving

\displaystyle \begin{align*} \int{\frac{\ln{(x + 1)}}{x + 1}\,dx} &= \int{\frac{\ln{(u)}}{u}\,du} \\ &= \int{\ln{(u)}\cdot \frac{1}{u}\,du} \end{align*}

Now notice that \displaystyle \begin{align*} \frac{1}{u} \end{align*} is the derivative of \displaystyle \begin{align*} \ln{(u)} \end{align*}, so a substitution of the form \displaystyle \begin{align*} v = \ln{(u)} \implies dv = \frac{1}{u}\,du \end{align*} is appropriate, giving

\displaystyle \begin{align*} \int{\ln{(u)}\cdot \frac{1}{u}\,du} = \int{ v\,dv} \end{align*}

Surely you can integrate THAT.