-x-1-x1-x ex2 d x is equal toA- logx ex-1-x ex-1-1-e xx-CB- logx-1-xex x a x-1-1-x x x-CC- log1- xex - xe x -1-1- xe x - CD- None of these

The correct option is A log ( xe^x/1+xe^x )+1/1+ex^x+C1+xe^x=t ⇒ (x+1)e^x dx = dt ∴∫(x+1)/x(x+xe^x)^2dx = ∫(x+1)e^x dx/xe^x(1+xe^x)^2 =∫dt/(t 1)t^2 = ∫ ( 1/ ...

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

Byju's Answer

Standard XII

Mathematics

Standard Formulae - 2

∫x+1/x1+x ex2...

Question

$\int \frac{(x + 1)}{x (1 + x e^{x})^{2}} d x$ is equal to

l o g (\frac{x e^{x}}{1 + x e^{x}}) + \frac{1}{1 + e x^{x}} + C

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

l o g (\frac{x}{1 + x e^{x}}) + \frac{1}{1 + x e^{x}} + C

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

l o g (\frac{1 + x e^{x}}{x e^{x}}) + \frac{1}{1 + x e^{x}} + C

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

None of these

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

Open in App

Solution

The correct option is A $l o g (\frac{x e^{x}}{1 + x e^{x}}) + \frac{1}{1 + e x^{x}} + C$
$1 + x e^{x} = t \Rightarrow (x + 1) e^{x} d x = d t ∴ \int \frac{(x + 1)}{x (x + x e^{x})^{2}} d x = \int \frac{(x + 1) e^{x} d x}{x e^{x} (1 + x e^{x})^{2}} = \int \frac{d t}{(t - 1) t^{2}} = \int (- \frac{1}{t^{2}} - \frac{1}{t} + \frac{1}{t - 1}) d t$ (by partial fractions)
$= \frac{1}{t} - l o g t + l o g (t - 1) + C = \frac{1}{t} + l o g (\frac{t - 1}{t}) + C = \frac{1}{1 + x e^{x}} + l o g (\frac{x e^{x}}{1 + x e^{x}}) + C$

Suggest Corrections

Similar questions

Q. The integral

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

is equal to:

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

is equal to

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

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

is equal to

(a) − xe^x + C

(b) xe^x + C

(c) − xe^−x + C

(d) xe^−x + C

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

(a) 2 log_e cos (xe^x) + C

(b) sec (xe^x) + C

(c) tan (xe^x) + C

(d) tan (x + e^x) + C

Join BYJU'S Learning Program

∫(x+1)x(1+xex)2dx is equal to

The correct option is A log(xex1+xex)+11+exx+C1+xex=t⇒(x+1)exdx=dt∴∫(x+1)x(x+xex)2dx=∫(x+1)exdxxex(1+xex)2=∫dt(t−1)t2=∫(−1t2−1t+1t−1)dt (by partial fractions) =1t−log t+log(t−1)+C=1t+log(t−1t)+C=11+xex+log(xex1+xex)+C

$\int \frac{(x + 1)}{x (1 + x e^{x})^{2}} d x$ is equal to