What is the integral of (36)/(1+16x^2) ?

The integral of (36)/(1+16x^2) is 9arctan(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (36)/(1+16x^2)

\int \frac{36}{1 + 16 x ^{2}} d x

9 arctan (4 x) + C

Solution steps

\int \frac{36}{1 + 16 x ^{2}} d x

Take the constant out:

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

= 36 \cdot \int \frac{1}{1 + 16 x ^{2}} d x

Apply Integral Substitution

= 36 \cdot \int \frac{1}{4 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 36 \cdot \frac{1}{4} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 36 \cdot \frac{1}{4} arctan (u)

Substitute back

u = 4 x

= 36 \cdot \frac{1}{4} arctan (4 x)

Simplify

36 \cdot \frac{1}{4} arctan (4 x) : 9 arctan (4 x)

= 9 arctan (4 x)

Add a constant to the solution

= 9 arctan (4 x) + C

\int \frac{3 x ^{2}}{b ^{5 x^{3}}} d x

\int \frac{2 x + 3}{x ^{2} + 3 x} d x

\int (1 - 3 x)^{5} d x

\int sin (x) cos^{8} (x) d x

\int 3 x^{2} x d x

What is the integral of (36)/(1+16x^2) ?
The integral of (36)/(1+16x^2) is 9arctan(4x)+C