The correct option is A e^xx+1+C xe^x( x + 1)^2dx xe^x. 1( x + 1)^2dx xe^x dx( x + 1)^2 [ ( xe^2 + e^x) × dx( 1 + x)^2]dx xe^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration by Parts

Solve ∫xex/...

Question

Solve $\int \frac{x e^{x}}{{(x + 1)}^{2}} d x$

\frac{e^{x}}{x + 1} + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{x}{(x + 1)^{2}} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{x} (x + 1) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x (x + 1)^{2} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{e^{x}}{x + 1} + C$
$\int \frac{x e^{x}}{{(x + 1)}^{2}} d x$
$\int x e^{x} . \frac{1}{{(x + 1)}^{2}} d x$
$x e^{x} \int \frac{d x}{{(x + 1)}^{2}} - \int [(x e^{2} + e^{x}) \times \int \frac{d x}{{(1 + x)}^{2}}] d x$
$x e^{x} (\frac{- 1}{x + 1}) - \int \frac{- (x e^{x} + e^{x})}{x + 1} d x$
$\frac{- x e^{x}}{x + 1} + \int e^{x} d x$
$\frac{- x e^{x}}{x + 1} + e^{x} + c$
$\frac{- x e^{x} + e^{x} (x + 1)}{x + 1} + c$
$\frac{e^{x}}{x + 1} + c$

Suggest Corrections

Similar questions

\int \frac{2}{{(e^{x} + e^{- x})}^{2}} d x

\frac{- e^{- x}}{e^{x} + e^{- x}} + C

- \frac{1}{e^{x} + e^{- x}} + C

\frac{- 1}{{(e^{x} + 1)}^{2}} + C

\frac{1}{e^{x} - e^{- x}} + C

\int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x

- e^{x} \tan \frac{x}{2} + C

- \frac{1}{2} e^{x} \tan \frac{x}{2} + C

- \frac{1}{2} e^{x} \cot \frac{x}{2} + C

Solution of differential equation $\frac{d y}{d x} + \frac{tan y}{x} = x e^{x} sec y$ is

$\int \frac{x e^{x}}{(1 + x)^{2}} d x =$

f : R \to \{x \in R : x < 1\}

f (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}

\frac{1}{2} \log \frac{1 + x}{1 - x}

\frac{1}{2} \log \frac{2 + x}{2 - x}

\frac{1}{2} \log \frac{1 - x}{1 + x}

Join BYJU'S Learning Program