The correct option is C e^x/(1 + x) + c ∫xe^x(1 + x)^2 dx = ∫ e^x (x + 1 1)(x + 1)^2 dx = ∫ e^x [ 1x + 1 + 1(x + 1)^2 ] dx = e^xx + 1 + c ...

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

Division of a Polynomial by a Polynomial

∫xex/1 + x2 d...

Question

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

\frac{e^{x}}{(1 + x)} + c

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

\frac{x e^{x}}{(1 + x)} + c

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

\frac{e^{x}}{(1 + x)^{3}} + c

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

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

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

Open in App

Solution

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

Suggest Corrections

Similar questions

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

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

\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C

\int e^{x} [\frac{x^{3} + x + 1}{(1 + x^{2})^{3 / 2}}] d x

\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 \frac{x e^{x}}{\sqrt{1 + e^{x}}} d x = f (x) \sqrt{1 + e^{x}} - 2 log g (x) + C

Join BYJU'S Learning Program