What is the integral of (8-128x)/(sqrt(1-64x^2)) ?

The integral of (8-128x)/(sqrt(1-64x^2)) is arcsin(8x)+2sqrt(1-64x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8-128x)/(sqrt(1-64x^2))

\int \frac{8 - 128 x}{1 - 64 x ^{2}} d x

arcsin (8 x) + 2 1 - 64 x^{2} + C

Solution steps

\int \frac{8 - 128 x}{1 - 64 x ^{2}} d x

Expand

\frac{8 - 128 x}{1 - 64 x ^{2}} : \frac{8}{1 - 64 x ^{2}} - \frac{128 x}{1 - 64 x ^{2}}

= \int \frac{8}{1 - 64 x ^{2}} - \frac{128 x}{1 - 64 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{8}{1 - 64 x ^{2}} d x - \int \frac{128 x}{1 - 64 x ^{2}} d x

\int \frac{8}{1 - 64 x ^{2}} d x = arcsin (8 x)

\int \frac{128 x}{1 - 64 x ^{2}} d x = - 2 1 - 64 x^{2}

= arcsin (8 x) - (- 2 1 - 64 x^{2})

Simplify

= arcsin (8 x) + 2 1 - 64 x^{2}

Add a constant to the solution

= arcsin (8 x) + 2 1 - 64 x^{2} + C

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

\int 12 x e^{- x^{2}} d x

\int \frac{7}{x ^{2} + 25} d x

\frac{d}{d x} (4 x^{\frac{2}{3}} + x^{\frac{5}{3}})

f (x) = x^{3} - 2 x, at x = 2

What is the integral of (8-128x)/(sqrt(1-64x^2)) ?
The integral of (8-128x)/(sqrt(1-64x^2)) is arcsin(8x)+2sqrt(1-64x^2)+C