Evaluate the definite integral. ∫0e−1x3x2+1dx

Question

Evaluate the definite integral.  ∫0e−1x3x2+1dx

Parminquale · Accepted Answer

Step 1  The given integral is ∫0e−1x3x2+1dx  Step 2  Let u=x2+1.  Then du =2xdx.  ∫0e−1x3x2+1dx=∫0e−1x2(x)x2+1dx  =∫1eu−1u(du2)  =12∫1e(1−1u)du  =12[u−ln⁡u]1e  =12[(e−ln⁡e)−(1−ln⁡1)]  =12(e−1−1)  =e−22  Hence, the value of the integral ∫0e−1x3x2+1dx is e−22.

Pulad1971 · Accepted Answer

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 (x) d x = F (x) ∣_{a}^{b} = F (b) - F (a)

Step 2: In this case,

f (x) = \frac{x^{3}}{x^{2} + 1}

. Find its integral.

\frac{x^{2}}{2} - \frac{\ln (x^{2} + 1)}{2} ∣_{0}^{\sqrt{e - 1}}

Step 3: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into

F (\sqrt{e - 1}) - F (0) :

(\frac{{\sqrt{e - 1}}^{2}}{2} - \frac{\ln ({\sqrt{e - 1}}^{2} + 1)}{2}) - (\frac{0^{2}}{2} - \frac{\ln (0^{2} + 1)}{2})

Step 4: Simplify.

\frac{e - 1}{2} - \frac{1}{2}

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

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