What is the integral of 3/(16x^2+1) ?

The integral of 3/(16x^2+1) is 3/4 arctan(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3/(16x^2+1)

\int \frac{3}{16 x ^{2} + 1} d x

\frac{3}{4} arctan (4 x) + C

Solution steps

\int \frac{3}{16 x ^{2} + 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{16 x ^{2} + 1} d x

Apply Integral Substitution

= 3 \cdot \int \frac{1}{4 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 3 \cdot \frac{1}{4} arctan (u)

Substitute back

u = 4 x

= 3 \cdot \frac{1}{4} arctan (4 x)

Simplify

3 \cdot \frac{1}{4} arctan (4 x) : \frac{3}{4} arctan (4 x)

= \frac{3}{4} arctan (4 x)

Add a constant to the solution

= \frac{3}{4} arctan (4 x) + C

\int ln (x) (3 x^{2} + 7) d x

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

\int \frac{e ^{9 x}}{e ^{9 x} + 7} d x

\int 5 e^{x} + \frac{4}{x} d x

\int (e^{9 x}) d x

What is the integral of 3/(16x^2+1) ?
The integral of 3/(16x^2+1) is 3/4 arctan(4x)+C