in Calculus edited by
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$\int \frac{x^3}{\sqrt{1+x^2}}.dx$
in Calculus edited by
by
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4 Comments

nope

answer is $\frac{1}{3}(1+x^2)^{3/2}-(1+x^2)^{1/2}+C$
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I am getting

$x^2 \sqrt{1+x^2} -\frac{2}{3}\sqrt[3]{1+x^2}$
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edited by
aditi put t=tan(inverse)x in my answer.you will get the same answer

use

sec(tan−1(x))=√(x^2+1)
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1 Answer

1 vote
1 vote

use 

sec(tan−1(x))=√x^2+1

prove of this inverse formula is here:https://socratic.org/questions/how-do-you-simplify-sec-tan-1-x

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