Integral using Substitution (Complex Exponential Functions)

I = \int e^{2 x^{3} - 15 x^{2} - 24 x - 6} (x^{2} - 5 x - 4) d x

Put

u = 2 x^{3} - 15 x^{2} - 24 x - 6

, then

d u = (6 x^{2} - 30 x - 24) d x

Or,

\frac{1}{6} d u = (x^{2} - 5 x - 4) d x

So,

I = \int e^{u} (\frac{1}{6}) d u

Or,

I = \frac{1}{6} \int e^{u} d u

Or,

I = \frac{1}{6} e^{u} + C

Or,

I = \frac{1}{6} e^{2 x^{3} - 15 x^{2} - 24 x - 6} + C

Algebra Analytic Geometry Differential Calculus Integral Calculus Home

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