What is the integral of 1/((2x+1)^2+4) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((2x+1)^2+4)

\int \frac{1}{( 2 x + 1 ) ^{2} + 4} d x

\frac{1}{4} arctan (\frac{1}{2} (2 x + 1)) + C

Solution steps

\int \frac{1}{( 2 x + 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 4 )} d u

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

Take the constant out:

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{2 x + 1}{2})

Simplify

\frac{1}{2} \cdot \frac{1}{2} arctan (\frac{2 x + 1}{2}) : \frac{1}{4} arctan (\frac{1}{2} (2 x + 1))

= \frac{1}{4} arctan (\frac{1}{2} (2 x + 1))

Add a constant to the solution

= \frac{1}{4} arctan (\frac{1}{2} (2 x + 1)) + C

\int \frac{( x - 1 )}{81 - 4 x ^{2}} d x

\int \frac{1}{1 + 9 cos ( x )} d x

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

\int \frac{5 x ^{2}}{( x - 3 ) ^{3}} d x

\int - a e^{- a x} d x

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