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$\int \frac{\mathrm{x}^3}{\sqrt{1+\mathrm{x}^2}} \mathrm{dx}=\mathrm{a}\left(1+\mathrm{x}^2\right)^{\frac{3}{2}}+\mathrm{b} \sqrt{1+\mathrm{x}^2}+\mathrm{c}$, (where $\mathrm{c}$ is constant of integration) then find the value of $a+b$
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Verified Answer
The correct answer is:
$\frac{-2}{3}$
Let $I=\int \frac{x^3}{\sqrt{1+x^2}} d x=\int \frac{x^2 x}{\sqrt{1+x^2}} d x$
Put $\sqrt{1+\mathrm{x}^2}=\mathrm{t} \Rightarrow 1+\mathrm{x}^2=\mathrm{t}^2 \quad \Rightarrow 2 \mathrm{xdx}=2 \mathrm{t} d \mathrm{t}$
$$
\therefore \mathrm{I}=\int \frac{\left(\mathrm{t}^2-1\right) \mathrm{t} \mathrm{dt}}{\mathrm{t}}=\int\left(\mathrm{t}^2-1\right) \mathrm{dt}=\frac{\mathrm{t}^3}{3}-\mathrm{t}+\mathrm{c}=\frac{\left(1+\mathrm{x}^2\right)^{\frac{3}{2}}}{3}-\sqrt{1-\mathrm{x}^2}+\mathrm{c}
$$
Comparing with given data, $\mathrm{a}=\frac{1}{3}, \mathrm{~b}=-1 \Rightarrow \mathrm{a}+\mathrm{b}=\frac{-2}{3}$
Put $\sqrt{1+\mathrm{x}^2}=\mathrm{t} \Rightarrow 1+\mathrm{x}^2=\mathrm{t}^2 \quad \Rightarrow 2 \mathrm{xdx}=2 \mathrm{t} d \mathrm{t}$
$$
\therefore \mathrm{I}=\int \frac{\left(\mathrm{t}^2-1\right) \mathrm{t} \mathrm{dt}}{\mathrm{t}}=\int\left(\mathrm{t}^2-1\right) \mathrm{dt}=\frac{\mathrm{t}^3}{3}-\mathrm{t}+\mathrm{c}=\frac{\left(1+\mathrm{x}^2\right)^{\frac{3}{2}}}{3}-\sqrt{1-\mathrm{x}^2}+\mathrm{c}
$$
Comparing with given data, $\mathrm{a}=\frac{1}{3}, \mathrm{~b}=-1 \Rightarrow \mathrm{a}+\mathrm{b}=\frac{-2}{3}$
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