What is the integral of e^{3x}-1/((x+4)^2) ?

The integral of e^{3x}-1/((x+4)^2) is 1/3 e^{3x}+1/(x+4)+C

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integral of e^{3x}-1/((x+4)^2)

\int e^{3 x} - \frac{1}{( x + 4 ) ^{2}} d x

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

Solution steps

\int e^{3 x} - \frac{1}{( x + 4 ) ^{2}} 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 e^{3 x} d x - \int \frac{1}{( x + 4 ) ^{2}} d x

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

\int \frac{1}{( x + 4 ) ^{2}} d x = - \frac{1}{x + 4}

= \frac{1}{3} e^{3 x} - (- \frac{1}{x + 4})

Simplify

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

Add a constant to the solution

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

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What is the integral of e^{3x}-1/((x+4)^2) ?
The integral of e^{3x}-1/((x+4)^2) is 1/3 e^{3x}+1/(x+4)+C