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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \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 \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} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

e^{- t} cos (4 t) + \frac{t}{5}

x \to 2 lim (sin (x - 2))

\frac{\partial}{\partial x} (50 - x^{2} - y^{2})

\frac{d}{d x} (x e^{x} + 1)

n = 1 \sum \infty 1 - (\frac{1}{2})^{n}

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