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

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- Aug 1st 2013, 06:32 AMsrirahulanIntegration
$\displaystyle \int[\frac{ln(x+1)}{x+1}]$ 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 AMJJacquelinRe: 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 AMPlatoRe: Integration
- Aug 1st 2013, 08:17 PMibduttRe: Integration
- Aug 1st 2013, 08:59 PMProve ItRe: Integration
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 $\displaystyle \displaystyle \begin{align*} u = x + 1 \implies du = dx \end{align*}$ would clean things up a bit, giving

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

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

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

Surely you can integrate THAT.