What is the integral of 1/(sqrt(4x^2+4x+17)) ?

The integral of 1/(sqrt(4x^2+4x+17)) is 1/2 ln((|2x+1+sqrt(4x^2+4x+17)|)/4)+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+17))

\int \frac{1}{4 x ^{2} + 4 x + 17} d x

\frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4}) + C

Solution steps

\int \frac{1}{4 x ^{2} + 4 x + 17} d x

Complete the square

4 x^{2} + 4 x + 17 : 4 (x + \frac{1}{2})^{2} + 16

= \int \frac{1}{4 ( x + \frac{1}{2} ) ^{2} + 16} 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 Trigonometric Substitution

: \int sec (v) d v

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

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

\frac{1}{2} ln tan (arctan (\frac{1}{2} (x + \frac{1}{2}))) + sec (arctan (\frac{1}{2} (x + \frac{1}{2}))) : \frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4})

= \frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4})

Add a constant to the solution

= \frac{1}{2} ln (\frac{2 x + 1 + 4 x ^{2} + 4 x + 17}{4}) + C

\int sin^{4} (x) \cdot cos^{3} (x) d x

\int \frac{ln ( 1 - x )}{e ^{2} 6 1 - x} d x

\int \frac{x ^{4} + 2 x + 1}{( x ^{2} + 1 ) ( x ^{2} + 9 )} d x

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

\int x^{2} (32 + 4 x^{3})^{6} d x

What is the integral of 1/(sqrt(4x^2+4x+17)) ?
The integral of 1/(sqrt(4x^2+4x+17)) is 1/2 ln((|2x+1+sqrt(4x^2+4x+17)|)/4)+C