What is the integral of+(4-32x)/(sqrt(1-16x^2)) ?

The integral of+(4-32x)/(sqrt(1-16x^2)) is arcsin(4x)+2sqrt(1-16x^2)+C

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integral of+(4-32x)/(sqrt(1-16x^2))

\int + \frac{4 - 32 x}{1 - 16 x ^{2}} d x

arcsin (4 x) + 2 1 - 16 x^{2} + C

Solution steps

\int \frac{4 - 32 x}{1 - 16 x ^{2}} d x

Expand

\frac{4 - 32 x}{1 - 16 x ^{2}} : \frac{4}{1 - 16 x ^{2}} - \frac{32 x}{1 - 16 x ^{2}}

= \int \frac{4}{1 - 16 x ^{2}} - \frac{32 x}{1 - 16 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}{1 - 16 x ^{2}} d x - \int \frac{32 x}{1 - 16 x ^{2}} d x

\int \frac{4}{1 - 16 x ^{2}} d x = arcsin (4 x)

\int \frac{32 x}{1 - 16 x ^{2}} d x = - 2 1 - 16 x^{2}

= arcsin (4 x) - (- 2 1 - 16 x^{2})

Simplify

= arcsin (4 x) + 2 1 - 16 x^{2}

Add a constant to the solution

= arcsin (4 x) + 2 1 - 16 x^{2} + C

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

x^{2} \frac{d y}{d x} + x y = 1

\frac{d}{d x} (x^{\frac{2}{3}} (2 x - x^{2}))

\frac{\partial ^{3}}{\partial x \partial y \partial z} (e^{x y z^{6}})

\frac{\partial}{\partial y} (3 e^{x} ln (y))

What is the integral of+(4-32x)/(sqrt(1-16x^2)) ?
The integral of+(4-32x)/(sqrt(1-16x^2)) is arcsin(4x)+2sqrt(1-16x^2)+C