Integral using Substitution (Simple Exponential Functions)

I = \int e^{\frac{3}{4} x + 1} d x

Put

u = \frac{3}{4} x + 1, then

d u = (\frac{3}{4}) d x

Or,

(\frac{4}{3}) d u = d x

So,

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

Or,

I = \frac{4}{3} \int e^{u} d u

Or,

I = \frac{4}{3} e^{u} + C

Or,

I = \frac{4}{3} e^{\frac{3}{4} x + 1} + C

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