What is the integral of 1/(sqrt(4x^2+4x+5)) ?

The integral of 1/(sqrt(4x^2+4x+5)) is 1/2 ln|(sqrt(4x^2+4x+5))/2+x+1/2 |+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(4x^2+4x+5))

\int \frac{1}{4 x ^{2} + 4 x + 5} d x

\frac{1}{2} ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2} + C

Solution steps

\int \frac{1}{4 x ^{2} + 4 x + 5} d x

Complete the square

4 x^{2} + 4 x + 5 : 4 (x + \frac{1}{2})^{2} + 4

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

Apply u-substitution

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

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

Use the common integral:

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

= \frac{1}{2} ln u^{2} + 1 + u

Substitute back

u = x + \frac{1}{2}

= \frac{1}{2} ln (x + \frac{1}{2})^{2} + 1 + x + \frac{1}{2}

Simplify

\frac{1}{2} ln (x + \frac{1}{2})^{2} + 1 + x + \frac{1}{2} : \frac{1}{2} ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2}

= \frac{1}{2} ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2}

Add a constant to the solution

= \frac{1}{2} ln \frac{4 x ^{2} + 4 x + 5}{2} + x + \frac{1}{2} + C

\int \frac{sinh ( 4 x )}{cosh ( 4 x )} d x

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

\int \frac{1}{x ^{- 9}} d x

\int x ln (x) d x

\int - \frac{2 y cos ( \frac{π y}{2} )}{π} d y

What is the integral of 1/(sqrt(4x^2+4x+5)) ?
The integral of 1/(sqrt(4x^2+4x+5)) is 1/2 ln|(sqrt(4x^2+4x+5))/2+x+1/2 |+C