- e -x- x x - x dx equalsA- e -x- x -2 - x -1- 2 e - x - x - x -1- CC- e - x - x - x - CD- e -x- x - x -1- C

The correct option is B 2e^√(x)[x √(x)+1]+C∫e^√(x)/√(x)(x+√(x))dx;; Put x=t^2; dx =2t dt =∫ e^t(t^2+t)dt=e^t(At^2+Bt+C) Diffrentiate both the sides e^t(t^2+t)= ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫ e √x/√ x x ...

Question

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) d x$ equals

2 e^{\sqrt{x}} [x - \sqrt{x} + 1] + C

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e^{\sqrt{x}} [x - 2 \sqrt{x} + 1]

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e^{\sqrt{x}} [x + \sqrt{x}] + C

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e^{\sqrt{x}} [x + \sqrt{x} + 1] + C

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Solution

The correct option is A $2 e^{\sqrt{x}} [x - \sqrt{x} + 1] + C$
$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) d x$ ;; Put $x = t^{2}; d x = 2 t d t$
$= \int e^{t} (t^{2} + t) d t = e^{t} (A t^{2} + B t + C)$
Diffrentiate both the sides
$e^{t} (t^{2} + t) = e^{t} (2 A t + B) + (A t^{2} + B t + C) e^{t}$
On comparing coefficient we get
A = 1 ; B = – 1 ; C = 1

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∫e√x√x(x+√x)dx equals

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$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) d x$ equals