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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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