Solve the indefinite integral. ∫23x3−2xx−1dx

Question

Solve the indefinite integral.  ∫23x3−2xx−1dx

Sue Leahy · Accepted Answer

Step 1  Some integration formulas:  ∫xndx=xn+1n+1+C  ∫1xdx=log⁡x+C  Given integral is I=∫23x3−2xx−1dx  Let us assume:  x−1=u  dx=du   Transformed integral is given as:  I=∫12(u+1)3−2(u+1)udu  =∫12(u2+3u−1u+1)du  Step 2  Next, integrate terms in pervious steps to get simplified form:   I=[u33+3u22−ln⁡(u)+u]12  =[83+6−ln⁡(2)+2−13−32+ln⁡(1)−1]  =[73+7−32−ln⁡(2)]  =476−ln⁡(2)  =7.14  Hence ∫23x3−2xx−1dx=7.14

John Twitchell · 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} - 2 x}{x - 1}

. Find its integral.

\frac{x^{3}}{3} + \frac{x^{2}}{2} - x - \ln (x - 1) ∣_{2}^{3}

Step 3: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(3)−F(2):

(\frac{3^{3}}{3} + \frac{3^{2}}{2} - 3 - \ln (3 - 1)) - (\frac{2^{3}}{3} + \frac{2^{2}}{2} - 2 - \ln (2 - 1))

Step 4: Simplify.

\frac{47}{6} - \ln 2

Solve the indefinite integral. \int_{2}^{3}\frac{x^{3}-2x}{x-1}dx

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