Integral using Substitution (Polynomials and Rationals)

I = \int \frac{8 x + 6}{{(16 x^{2} + 24 x - 1)}^{3}} d x

Put u = 16 x^{2} + 24 x - 1, then

d u = (32 x + 24) d x

or, (\frac{1}{4}) d u = (8 x + 6) d x

I = \int {(u)}^{-3} (\frac{1}{4}) d u

I = \frac{1}{4} \int {(u)}^{-3} d u

I = \frac{1}{4} \frac{{(u)}^{-2}}{-2} + C

I = \frac{1}{4} \frac{-1}{2 {(u)}^{2}} + C

I = \frac{-1}{8 {(16 x^{2} + 24 x - 1)}^{2}} + C

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