pro4ph5e4q2

2021-11-18

Evaluate the definite integral.
${\int }_{0}^{\sqrt{e-1}}\frac{{x}^{3}}{{x}^{2}+1}dx$

Parminquale

Step 1
The given integral is ${\int }_{0}^{\sqrt{e-1}}\frac{{x}^{3}}{{x}^{2}+1}dx$
Step 2
Let $u={x}^{2}+1$.
Then du =2xdx.
${\int }_{0}^{\sqrt{e-1}}\frac{{x}^{3}}{{x}^{2}+1}dx={\int }_{0}^{\sqrt{e-1}}\frac{{x}^{2}\left(x\right)}{{x}^{2}+1}dx$
$={\int }_{1}^{e}\frac{u-1}{u}\left(\frac{du}{2}\right)$
$=\frac{1}{2}{\int }_{1}^{e}\left(1-\frac{1}{u}\right)du$
$=\frac{1}{2}{\left[u-\mathrm{ln}u\right]}_{1}^{e}$
$=\frac{1}{2}\left[\left(e-\mathrm{ln}e\right)-\left(1-\mathrm{ln}1\right)\right]$
$=\frac{1}{2}\left(e-1-1\right)$
$=\frac{e-2}{2}$
Hence, the value of the integral .

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
${\int }_{a}^{b}f\left(x\right)dx=F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$
Step 2: In this case, $f\left(x\right)=\frac{{x}^{3}}{{x}^{2}+1}$. Find its integral.
$\frac{{x}^{2}}{2}-\frac{\mathrm{ln}\left({x}^{2}+1\right)}{2}{\mid }_{0}^{\sqrt{e-1}}$
Step 3: Since $F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$, expand the above into $F\left(\sqrt{e-1}\right)-F\left(0\right):$
$\left(\frac{{\sqrt{e-1}}^{2}}{2}-\frac{\mathrm{ln}\left({\sqrt{e-1}}^{2}+1\right)}{2}\right)-\left(\frac{{0}^{2}}{2}-\frac{\mathrm{ln}\left({0}^{2}+1\right)}{2}\right)$
Step 4: Simplify.
$\frac{e-1}{2}-\frac{1}{2}$

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