What is the integral of 1/(sqrt(2x^2+8x+10)) ?

The integral of 1/(sqrt(2x^2+8x+10)) is 1/(sqrt(2))ln|x+2+sqrt(x^2+4x+5)|+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(2x^2+8x+10))

\int \frac{1}{2 x ^{2} + 8 x + 10} d x

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

Solution steps

\int \frac{1}{2 x ^{2} + 8 x + 10} d x

Complete the square

2 x^{2} + 8 x + 10 : 2 (x + 2)^{2} + 2

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= \int \frac{sec ( v )}{2} 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 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{2}{2} (x + 2))) + sec (arctan (\frac{2}{2} (x + 2)))

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x ^{2} + 1}{( x - 2 ) ^{3}} d x

\int \frac{1}{w ^{2} w ^{2} + a ^{2}} d w

\int \frac{3}{2} cos (x) d x

\int x^{3} - 6 x^{2} + 9 x + 6 d x

\int \frac{14 x - 46}{( x - 2 ) ( x - 5 )} d x

What is the integral of 1/(sqrt(2x^2+8x+10)) ?
The integral of 1/(sqrt(2x^2+8x+10)) is 1/(sqrt(2))ln|x+2+sqrt(x^2+4x+5)|+C