What is the integral of 1/(9x^2+25) ?

The integral of 1/(9x^2+25) is 1/15 arctan(3/5 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(9x^2+25)

\int \frac{1}{9 x ^{2} + 25} d x

\frac{1}{15} arctan (\frac{3}{5} x) + C

Solution steps

\int \frac{1}{9 x ^{2} + 25} d x

Apply Integral Substitution

= \int \frac{1}{15 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{15} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{15} arctan (u)

Substitute back

u = \frac{3}{5} x

= \frac{1}{15} arctan (\frac{3}{5} x)

Add a constant to the solution

= \frac{1}{15} arctan (\frac{3}{5} x) + C

\int \frac{x}{x ^{2} - 20 x} d x

\int 0.25 x d x

\int \frac{3 x}{ln ( x )} d x

x e^{2 x}

\int \frac{3}{( 1 + 9 x ^{2} )} d x

What is the integral of 1/(9x^2+25) ?
The integral of 1/(9x^2+25) is 1/15 arctan(3/5 x)+C