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

The integral of 1/(sqrt(1-4^2)) is -i(sqrt(15)x)/(15)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(1-4^2))

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

- i \frac{15 x}{15} + C

Solution steps

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

Integral of a constant:

\int a d x = a x

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

Simplify

\frac{1}{1 - 4 ^{2}} x : - i \frac{15 x}{15}

= - i \frac{15 x}{15}

Add a constant to the solution

= - i \frac{15 x}{15} + C

\int_{- 4}^{4} (16 - x^{2}) d x

y^{^{'}} = \frac{x}{x - 1}, y (2) = 5

\int \frac{x}{x ^{2} - 8 x} d x

\frac{d}{d x} (5 x sin (x))

What is the integral of 1/(sqrt(1-4^2)) ?
The integral of 1/(sqrt(1-4^2)) is -i(sqrt(15)x)/(15)+C