Integral using Substitution (Simple Exponential Functions)

I = \int e^{\frac{3}{8} x + \frac{1}{3}} d x

Put

u = \frac{3}{8} x + \frac{1}{3}, then

d u = (\frac{3}{8}) d x

Or,

(\frac{8}{3}) d u = d x

So,

I = \int e^{u} (\frac{8}{3}) d u

Or,

I = \frac{8}{3} \int e^{u} d u

Or,

I = \frac{8}{3} e^{u} + C

Or,

I = \frac{8}{3} e^{\frac{3}{8} x + \frac{1}{3}} + C

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