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

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

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integral of (4x+3)/(sqrt(1-4x^2))

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

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

Solution steps

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

Expand

\frac{4 x + 3}{1 - 4 x ^{2}} : \frac{4 x}{1 - 4 x ^{2}} + \frac{3}{1 - 4 x ^{2}}

= \int \frac{4 x}{1 - 4 x ^{2}} + \frac{3}{1 - 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{4 x}{1 - 4 x ^{2}} d x + \int \frac{3}{1 - 4 x ^{2}} d x

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

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

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

Add a constant to the solution

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

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