∫x3dx1+x2 is equal to

I=∫x3dx1+x2=∫x⋅x2dx1+x2 Let t=1+x2⇒dtdx=x1+x2 Now, I=∫t2−1dt=t33−t+C=t3t2−3+C∴ I=131+x2x2−2+C

∫x3dx1+x2 is equal to

I=∫x3dx1+x2=∫x⋅x2dx1+x2 Let t=1+x2⇒dtdx=x1+x2 Now, I=∫t2−1dt=t33−t+C=t3t2−3+C∴ I=131+x2x2−2+C

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$\int \frac{x^{3} d x}{\sqrt{1 + x^{2}}} is equal to$

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$\frac{1}{3} \sqrt{1 + x^{2}} (2 + x^{2}) + C$

$\frac{1}{3} \sqrt{1 + x^{2}} (x^{2} - 1) + C$

$\frac{1}{3} {(1 + x^{2})}^{3 / 2} + C$

$\frac{1}{3} \sqrt{1 + x^{2}} (x^{2} - 2) + C$

answer is D.

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Detailed Solution

$I = \int \frac{x^{3} d x}{\sqrt{1 + x^{2}}} = \int \frac{x \cdot x^{2} d x}{\sqrt{1 + x^{2}}}$

$Let t = \sqrt{1 + x^{2}} \Rightarrow \frac{d t}{d x} = \frac{x}{\sqrt{1 + x^{2}}}$

$Now, I = \int (t^{2} - 1) d t = \frac{t^{3}}{3} - t + C = \frac{t}{3} (t^{2} - 3) + C$

$∴ I = \frac{1}{3} \sqrt{1 + x^{2}} (x^{2} - 2) + C$

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$\int \frac{x^{3} d x}{\sqrt{1 + x^{2}}} is equal to$