-x-1-x-1-xe-x-2dx is equal tolog - x-1-xe-x -1-1-xe-x-CNone of theselog - 1-xe-x-xe-x -1-1-xe-x-Clog - xe-x-1-xe-x -1-1-ex-x-C

The correct option is D 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/ ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫(x+1)/x(1+xe...

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

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

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

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None of these

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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)
$= \frac{1}{t} - l o g t + l o g (t - 1) + C = \frac{1}{t} + l o g (\frac{t - 1}{t}) + C = \frac{1}{1 + x e^{x}} + l o g (\frac{x e^{x}}{1 + x e^{x}}) + C$

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Similar questions

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

is equal to

Q. The integral

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

is equal to:

\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

Q. Let

y = x e^{x}

then prove that

x \frac{d y}{d x} = (1 + x) e^{x}

$l i m x \to 0$ $\frac{x (e^{x} - 1)}{1 - c o s x}$ =

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∫(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