Integral using Substitution (Polynomials and Rationals)

I = \int \frac{10 x^{2} + 4 x - 6}{{(40 x^{3} + 24 x^{2} - 72 x - 3)}^{3}} d x

Put u = 40 x^{3} + 24 x^{2} - 72 x - 3, then

d u = (120 x^{2} + 48 x - 72) d x

or, (\frac{1}{12}) d u = (10 x^{2} + 4 x - 6) d x

I = \int {(u)}^{-3} (\frac{1}{12}) d u

I = \frac{1}{12} \int {(u)}^{-3} d u

I = \frac{1}{12} \frac{{(u)}^{-2}}{-2} + C

I = \frac{1}{12} \frac{-1}{2 {(u)}^{2}} + C

I = \frac{-1}{24 {(40 x^{3} + 24 x^{2} - 72 x - 3)}^{2}} + C

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Copyright © Dayal D. Purohit, Ph.D.(Mathematics)