What is the integral of (sqrt(25-16^2))/x ?

The integral of (sqrt(25-16^2))/x is sqrt(231)iln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(25-16^2))/x

\int \frac{25 - 1 6 ^{2}}{x} d x

231 i ln ∣ x ∣ + C

Solution steps

\int \frac{25 - 1 6 ^{2}}{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 25 - 1 6^{2} \cdot \int \frac{1}{x} d x

Simplify

25 - 1 6^{2} : 231 i

= 231 i \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= 231 i ln ∣ x ∣

Add a constant to the solution

= 231 i ln ∣ x ∣ + C

\int \frac{2 x + 1}{( x + 5 ) ( x - 2 )} d x

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

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

\int (x^{8} - x^{9}) ln (x) d x

\int - 2 x^{6} d x

What is the integral of (sqrt(25-16^2))/x ?
The integral of (sqrt(25-16^2))/x is sqrt(231)iln|x|+C