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

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

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

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

\frac{1}{2} arcsin (\frac{2}{3} x) - \frac{1}{2} 9 - 4 x^{2} + C

Solution steps

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

Expand

\frac{1 + 2 x}{9 - 4 x ^{2}} : \frac{1}{9 - 4 x ^{2}} + \frac{2 x}{9 - 4 x ^{2}}

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

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

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

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

Add a constant to the solution

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

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