The correct option is B 2 e^ √(x)( √(x) 1 ) + c I=∫ #160;e^√( x ) #160;dx Put √( x ) =t #160; 1 / 2 x ^ 1/2 dx=dt ⇒ dx=2tdt So, ...

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Standard XII

Mathematics

Special Integrals - 1

∫ e√xdx=

Question

$\int e^{\sqrt{x}} d x =$

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

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

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e^{x} (x - 1) + c

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

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Solution

The correct option is B $2 e^{\sqrt{x}} (\sqrt{x} - 1) + c$
$I = \int e^{\sqrt{x}} d x$
Put $\sqrt{x} = t$
$\frac{1}{2} x^{- 1 / 2} d x = d t$
$\Rightarrow d x = 2 t d t$
So, $I = 2 \int t e^{t} d t$
$= 2 [t e^{t} - \int e^{t} d t]$
$= 2 t e^{t} - 2 e^{t} + C$
$= 2 e^{\sqrt{x}} (\sqrt{x} - 1) + c$

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∫e√xdx=

The correct option is B 2e√x(√x−1)+cI=∫e√xdxPut √x=t12x−1/2dx=dt⇒dx=2tdtSo, I=2∫tetdt=2[tet−∫etdt]=2tet−2et+C=2e√x(√x−1)+c

$\int e^{\sqrt{x}} d x =$

The correct option is B $2 e^{\sqrt{x}} (\sqrt{x} - 1) + c$
$I = \int e^{\sqrt{x}} d x$
Put $\sqrt{x} = t$
$\frac{1}{2} x^{- 1 / 2} d x = d t$
$\Rightarrow d x = 2 t d t$
So, $I = 2 \int t e^{t} d t$
$= 2 [t e^{t} - \int e^{t} d t]$
$= 2 t e^{t} - 2 e^{t} + C$
$= 2 e^{\sqrt{x}} (\sqrt{x} - 1) + c$