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If $\int \frac{x^3}{\sqrt{1+x^2}} d x=A\left(1+x^2\right)^{\frac{3}{2}}+B\left(1+x^2\right)^{\frac{1}{2}}+C$, then $\mathrm{A}+\mathrm{B}=$
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The correct answer is:
$-\frac{2}{3}$
$\int \frac{x^3}{\sqrt{1+x^2}} d x=\mathrm{A}\left(1+x^2\right)^{3 / 2}+\mathrm{B}\left(1+x^2\right)^{1 / 2}+c$ ...(i)
$\int \frac{x^3}{\sqrt{1+x^2}} d x=\int \frac{x^2 \cdot x}{\sqrt{1+x^2}} d x$
Let $1+x^2=t^2 \Rightarrow x d x=t d t$
$\therefore \int \frac{x^3}{\sqrt{1+x^2}} d x=\int \frac{\left(t^2-1\right)}{t} t d t$
$=\int\left(t^2-1\right) d t=\frac{t^3}{3}-t+c$
$\Rightarrow \int \frac{x^3}{\sqrt{1+x^2}} d x=\frac{1}{3}\left(1+x^2\right)^{3 / 2}-\left(1+x^2\right)^{1 / 2}+\mathrm{c}$ ...(ii)
From $\mathrm{eq}^{\mathrm{n}}$ (i) \& (ii), we get
$\begin{aligned} & A=\frac{1}{3}, B=-1 \\ & A+B=\frac{1}{3}-1=-\frac{2}{3}\end{aligned}$
$\int \frac{x^3}{\sqrt{1+x^2}} d x=\int \frac{x^2 \cdot x}{\sqrt{1+x^2}} d x$
Let $1+x^2=t^2 \Rightarrow x d x=t d t$
$\therefore \int \frac{x^3}{\sqrt{1+x^2}} d x=\int \frac{\left(t^2-1\right)}{t} t d t$
$=\int\left(t^2-1\right) d t=\frac{t^3}{3}-t+c$
$\Rightarrow \int \frac{x^3}{\sqrt{1+x^2}} d x=\frac{1}{3}\left(1+x^2\right)^{3 / 2}-\left(1+x^2\right)^{1 / 2}+\mathrm{c}$ ...(ii)
From $\mathrm{eq}^{\mathrm{n}}$ (i) \& (ii), we get
$\begin{aligned} & A=\frac{1}{3}, B=-1 \\ & A+B=\frac{1}{3}-1=-\frac{2}{3}\end{aligned}$
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