What is the integral of 1/(sqrt(4-9x^2))+e^{4x} ?

The integral of 1/(sqrt(4-9x^2))+e^{4x} is 1/3 arcsin(3/2 x)+1/4 e^{4x}+C

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integral of 1/(sqrt(4-9x^2))+e^{4x}

\int \frac{1}{4 - 9 x ^{2}} + e^{4 x} d x

\frac{1}{3} arcsin (\frac{3}{2} x) + \frac{1}{4} e^{4 x} + C

Solution steps

\int \frac{1}{4 - 9 x ^{2}} + e^{4 x} 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{1}{4 - 9 x ^{2}} d x + \int e^{4 x} d x

\int \frac{1}{4 - 9 x ^{2}} d x = \frac{1}{3} arcsin (\frac{3}{2} x)

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

= \frac{1}{3} arcsin (\frac{3}{2} x) + \frac{1}{4} e^{4 x}

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{3}{2} x) + \frac{1}{4} e^{4 x} + C

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

\int \frac{4}{9 - x ^{2}} d x

\int sin (a) x cos (a) x d x

\int 1 - v^{2} d v

\int (e^{t} - e^{- t}) d t

What is the integral of 1/(sqrt(4-9x^2))+e^{4x} ?
The integral of 1/(sqrt(4-9x^2))+e^{4x} is 1/3 arcsin(3/2 x)+1/4 e^{4x}+C