What is the integral of (12/5 e^{5x-3}+2)^2 ?

The integral of (12/5 e^{5x-3}+2)^2 is 72/125 e^{10x-6}+48/25 e^{5x-3}+4x+C

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integral of (12/5 e^{5x-3}+2)^2

\int (\frac{12}{5} e^{5 x - 3} + 2)^{2} d x

\frac{72}{125} e^{10 x - 6} + \frac{48}{25} e^{5 x - 3} + 4 x + C

Solution steps

\int (\frac{12}{5} e^{5 x - 3} + 2)^{2} d x

Expand

(\frac{12}{5} e^{5 x - 3} + 2)^{2} : \frac{144}{25} e^{10 x - 6} + \frac{48}{5} e^{5 x - 3} + 4

= \int \frac{144}{25} e^{10 x - 6} + \frac{48}{5} e^{5 x - 3} + 4 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{144}{25} e^{10 x - 6} d x + \int \frac{48}{5} e^{5 x - 3} d x + \int 4 d x

\int \frac{144}{25} e^{10 x - 6} d x = \frac{72}{125} e^{10 x - 6}

\int \frac{48}{5} e^{5 x - 3} d x = \frac{48}{25} e^{5 x - 3}

\int 4 d x = 4 x

= \frac{72}{125} e^{10 x - 6} + \frac{48}{25} e^{5 x - 3} + 4 x

Add a constant to the solution

= \frac{72}{125} e^{10 x - 6} + \frac{48}{25} e^{5 x - 3} + 4 x + C

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What is the integral of (12/5 e^{5x-3}+2)^2 ?
The integral of (12/5 e^{5x-3}+2)^2 is 72/125 e^{10x-6}+48/25 e^{5x-3}+4x+C