If int frac{x^3}{sqrt{1+x^2}} d x=A(1+x^2)^{f | vedclass

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Question

(B) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} d x$. Substitute $1+x^2 = t^2$,which implies $2x d x = 2t d t$ or $x d x = t d t$. Also,$x^2 = t^2 - 1$. Su

Vedclass · Accepted Answer

$-\frac{2}{3}$

Vedclass · Answer

(B) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} d x$. Substitute $1+x^2 = t^2$,which implies $2x d x = 2t d t$ or $x d x = t d t$. Also,$x^2 = t^2 - 1$. Substituting these into the integral: $I = \int \frac{(t^2-1) t d t}{t} = \int (t^2-1) d t$. Integrating with respect to $t$: $I = \frac{t^3}{3} - t + C$. Substituting $t = (1+x^2)^{1/2}$ back: $I = \frac{1}{3}(1+x^2)^{3/2} - (1+x^2)^{1/2} + C$. Comparing this with $A(1+x^2)^{3/2} + B(1+x^2)^{1/2} + C$,we get $A = \frac{1}{3}$ and $B = -1$. Therefore,$A+B = \frac{1}{3} - 1 = -\frac{2}{3}$.

If $\int \frac{x^3}{\sqrt{1+x^2}} d x=A(1+x^2)^{\frac{3}{2}}+B(1+x^2)^{\frac{1}{2}}+C$,then $A+B=$

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