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

pro4ph5e4q2

pro4ph5e4q2

Answered question

2021-11-18

Evaluate the definite integral.
0e1x3x2+1dx

Answer & Explanation

Parminquale

Parminquale

Beginner2021-11-19Added 17 answers

Step 1
The given integral is 0e1x3x2+1dx
Step 2
Let u=x2+1.
Then du =2xdx.
0e1x3x2+1dx=0e1x2(x)x2+1dx
=1eu1u(du2)
=121e(11u)du
=12[ulnu]1e
=12[(elne)(1ln1)]
=12(e11)
=e22
Hence, the value of the integral 0e1x3x2+1dx is e22.
Pulad1971

Pulad1971

Beginner2021-11-20Added 22 answers

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
abf(x)dx=F(x)ab=F(b)F(a)
Step 2: In this case, f(x)=x3x2+1. Find its integral.
x22ln(x2+1)20e1
Step 3: Since F(x)ab=F(b)F(a), expand the above into F(e1)F(0):
(e122ln(e12+1)2)(022ln(02+1)2)
Step 4: Simplify.
e1212

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