Integral using Substitution (Complex Exponential Functions)

I = \int e^{2 x^{3} - 12 x^{2} - 36 x - 9} (x^{2} - 4 x - 6) d x

Put

u = 2 x^{3} - 12 x^{2} - 36 x - 9

, then

d u = (6 x^{2} - 24 x - 36) d x

Or,

\frac{1}{6} d u = (x^{2} - 4 x - 6) 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} - 12 x^{2} - 36 x - 9} + C

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