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

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

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

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

Solution steps

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

Complete the square

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

= \int \frac{1}{( x + 1 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{4} (x + 1))) + sec (arctan (\frac{1}{4} (x + 1)))

Simplify

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

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

Add a constant to the solution

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

\int csc (1) d x

- \frac{1}{( x - 2 ) ^{2}}

y^{^{'}} + 5 y = 3 sin (e^{5 x})

\frac{\partial}{\partial z} (x z^{2})

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

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