-t -t-t-t2 d t is equal toA- 1-2 -2- t -1- t - -B- 2- t -1- t - -C- 2 -2- t -1- t - cD- 1-2- t -1- t V - c

The correct option is B 2 ( √(t) 1√(t))+c∫ ( t√(t)+√(t)t^2 )dt =∫ d1/√(t)dt+∫√(t)/t^2×√(t)√(t)dt = ∫1√(t)dt+∫1t√(t)dt ∫ t^ 12dt+∫ t^ 32dt= t^( 12+1)( 12+1)+ ...

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Byju's Answer

Standard XII

Physics

Antiderivative

∫t √t+√t/t2 d...

Question

$\int (\frac{t \sqrt{t} + \sqrt{t}}{t^{2}}) d t$ is equal to

\frac{1}{2 \sqrt{2}} (\sqrt{t} - \frac{1}{\sqrt{t}}) + c

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2 (\sqrt{t} - \frac{1}{\sqrt{t}}) + c

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2 \sqrt{2} (\sqrt{t} - \frac{1}{\sqrt{t}}) + c

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\frac{1}{2} (\sqrt{t} - \frac{1}{\sqrt{t}}) + c

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Solution

The correct option is B $2 (\sqrt{t} - \frac{1}{\sqrt{t}}) + c$
$\int (\frac{t \sqrt{t} + \sqrt{t}}{t^{2}}) d t = \int d \frac{1}{\sqrt{t}} d t + \int \frac{\sqrt{t}}{t^{2}} \times \frac{\sqrt{t}}{\sqrt{t}} d t = \int \frac{1}{\sqrt{t}} d t + \int \frac{1}{t \sqrt{t}} d t$
$\int t^{\frac{- 1}{2}} d t + \int t^{\frac{- 3}{2}} d t = \frac{t^{(} - \frac{1}{2} + 1)}{(- \frac{1}{2} + 1)} + \frac{t^{(} - \frac{3}{2} + 1)}{(- \frac{3}{2} + 1)}$ $= 2 \sqrt{t} - \frac{2}{\sqrt{t}} + c = 2 (\sqrt{t} - \frac{1}{\sqrt{t}}) + c$

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∫(t√t+√tt2)dt is equal to

The correct option is B 2(√t−1√t)+c∫(t√t+√tt2)dt=∫d1√tdt+∫√tt2×√t√tdt=∫1√tdt+∫1t√tdt ∫t−12dt+∫t−32dt=t(−12+1)(−12+1)+t(−32+1)(−32+1) =2√t−2√t+c=2(√t−1√t)+c

$\int (\frac{t \sqrt{t} + \sqrt{t}}{t^{2}}) d t$ is equal to