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

Evaluate the definite integral.
$$\displaystyle{\int_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}{\frac{{{x}^{{{3}}}}}{{{x}^{{{2}}}+{1}}}}{\left.{d}{x}\right.}$$

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Parminquale
Step 1
The given integral is $$\displaystyle{\int_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}{\frac{{{x}^{{{3}}}}}{{{x}^{{{2}}}+{1}}}}{\left.{d}{x}\right.}$$
Step 2
Let $$\displaystyle{u}={x}^{{{2}}}+{1}$$.
Then du =2xdx.
$$\displaystyle{\int_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}{\frac{{{x}^{{{3}}}}}{{{x}^{{{2}}}+{1}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}{\frac{{{x}^{{{2}}}{\left({x}\right)}}}{{{x}^{{{2}}}+{1}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{{1}}}^{{{e}}}}{\frac{{{u}-{1}}}{{{u}}}}{\left({\frac{{{d}{u}}}{{{2}}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\int_{{{1}}}^{{{e}}}}{\left({1}-{\frac{{{1}}}{{{u}}}}\right)}{d}{u}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{{\left[{u}-{\ln{{u}}}\right]}_{{{1}}}^{{{e}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{\left({e}-{\ln{{e}}}\right)}-{\left({1}-{\ln{{1}}}\right)}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left({e}-{1}-{1}\right)}$$
$$\displaystyle={\frac{{{e}-{2}}}{{{2}}}}$$
Hence, the value of the integral $$\displaystyle{\int_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}{\frac{{{x}^{{{3}}}}}{{{x}^{{{2}}}+{1}}}}{\left.{d}{x}\right.}\ {i}{s}\ {\frac{{{e}-{2}}}{{{2}}}}$$.
###### Not exactly what youâ€™re looking for?
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
$$\displaystyle{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
Step 2: In this case, $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{3}}}}}{{{x}^{{{2}}}+{1}}}}$$. Find its integral.
$$\displaystyle{\frac{{{x}^{{{2}}}}}{{{2}}}}-{\frac{{{\ln{{\left({x}^{{{2}}}+{1}\right)}}}}}{{{2}}}}{{\mid}_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}$$
Step 3: Since $$\displaystyle{F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$, expand the above into $$\displaystyle{F}{\left(\sqrt{{{e}-{1}}}\right)}-{F}{\left({0}\right)}:$$
$$\displaystyle{\left({\frac{{\sqrt{{{e}-{1}}}^{{{2}}}}}{{{2}}}}-{\frac{{{\ln{{\left(\sqrt{{{e}-{1}}}^{{{2}}}+{1}\right)}}}}}{{{2}}}}\right)}-{\left({\frac{{{0}^{{{2}}}}}{{{2}}}}-{\frac{{{\ln{{\left({0}^{{{2}}}+{1}\right)}}}}}{{{2}}}}\right)}$$
Step 4: Simplify.
$$\displaystyle{\frac{{{e}-{1}}}{{{2}}}}-{\frac{{{1}}}{{{2}}}}$$