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

The integral of 1/(sqrt(x^2+4x-5)) is ln(1/3 |sqrt(x^2+4x-5)+x+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(x^2+4x-5))

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

ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + C

Solution steps

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

Complete the square

x^{2} + 4 x - 5 : (x + 2)^{2} - 9

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 9} d u

= \int \frac{1}{u ^{2} - 9} 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 (arcsec (\frac{1}{3} (x + 2))) + sec (arcsec (\frac{1}{3} (x + 2)))

Simplify

ln tan (arcsec (\frac{1}{3} (x + 2))) + sec (arcsec (\frac{1}{3} (x + 2))) : ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2)

= ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2)

Add a constant to the solution

= ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + C

t^{2} y^{^{'}} + 2 t y = y^{3}

y^{^{'}} = \frac{2 sec ( y )}{( x - 2 ) ^{2}}

f (x) = \frac{t ^{2} + 3}{t ^{4} - 2 t + 4}

x \to 0 lim (x + 2)

\int z^{31} + z^{8} d z

What is the integral of 1/(sqrt(x^2+4x-5)) ?
The integral of 1/(sqrt(x^2+4x-5)) is ln(1/3 |sqrt(x^2+4x-5)+x+2|)+C