Use integration by parts to evaluate each function

Question

I tried to solve this task,but I couldn't do anyway...Please,If you know help me!Thank you!

Use integration by parts to evaluate each function $$\displaystyle \int x^3 \sqrt{1+x^2} \, dx$$

Answer 1

HINT:

$$\int x^3\sqrt{1+x^2}\space\text{d}x=$$

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Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$:

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$$\frac{1}{2}\int u\sqrt{1+u}\space\text{d}u=$$

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Substitute $s=u+1$ and $\text{d}s=\text{d}u$:

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$$\frac{1}{2}\int (s-1)\sqrt{s}\space\text{d}s=$$ $$\frac{1}{2}\int\left(s^{\frac{3}{2}}-\sqrt{s}\right)\space\text{d}s=$$ $$\frac{1}{2}\int s^{\frac{3}{2}}\space\text{d}s-\frac{1}{2}\int\sqrt{s}\space\text{d}s=$$ $$\frac{1}{2}\int s^{\frac{3}{2}}\space\text{d}s-\frac{1}{2}\int s^{\frac{1}{2}}\space\text{d}s$$

Answer 2

Substitution Method It is easiest to use substitution

let $1+x^2=t^2\implies 2x\ dx$ $=2t\ dt\ \ \forall \ \ \ t>0$, hence $$\int x^3\sqrt{1+x^2}\ dx=\int x^2\sqrt{1+x^2}(x\ dx)$$ $$=\int (t^2-1)\sqrt{t^2}\ (tdt)$$$$=\int (t^3-t)|t|\ dt$$ $$=\int (t^3-t)t\ dt$$$$=\int t^4\ dt-\int t^2\ dt$$ $$=\frac{t^5}{5}-\frac{t^3}{3}+C$$ $$=\frac{(1+x^2)^{5/2}}{5}-\frac{(1+x^2)^{3/2}}{3}+C$$

Answer 3

If you want to do it by parts and not by substitution, you would have to let $$u=x^2 \Rightarrow \frac{du}{dx}=2x$$ and $$\frac{dv}{dx}=x(1+x^2)^{\frac 12}\Rightarrow v=\frac 13(1+x^2)^{\frac 32}$$ Then $$I=\frac{x^2}{3}(1+x^2)^{\frac 32}-\int \frac 23 x(1+x^2)^{\frac 32} dx$$ $$=\frac{x^2}{3}(1+x^2)^{\frac 32}-\frac {2}{15}(1+x^2)^{\frac 52} +c$$