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

pro4ph5e4q2 2021-11-18 Answered
Evaluate the definite integral.
\(\displaystyle{\int_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}{\frac{{{x}^{{{3}}}}}{{{x}^{{{2}}}+{1}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Parminquale
Answered 2021-11-19 Author has 7705 answers
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 \(\displaystyle{\int_{{{0}}}^{{\sqrt{{{e}-{1}}}}}}{\frac{{{x}^{{{3}}}}}{{{x}^{{{2}}}+{1}}}}{\left.{d}{x}\right.}\ {i}{s}\ {\frac{{{e}-{2}}}{{{2}}}}\).
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Pulad1971
Answered 2021-11-20 Author has 9023 answers
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.
\(\displaystyle{\frac{{{e}-{1}}}{{{2}}}}-{\frac{{{1}}}{{{2}}}}\)
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