Integral using Substitution (Simple Exponential Functions)

I = \int e^{\frac{6}{5} x + \frac{1}{6}} d x

Put

u = \frac{6}{5} x + \frac{1}{6}, then

d u = (\frac{6}{5}) d x

Or,

(\frac{5}{6}) d u = d x

So,

I = \int e^{u} (\frac{5}{6}) d u

Or,

I = \frac{5}{6} \int e^{u} d u

Or,

I = \frac{5}{6} e^{u} + C

Or,

I = \frac{5}{6} e^{\frac{6}{5} x + \frac{1}{6}} + C

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