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

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

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

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

\frac{1}{3} arcsec (\frac{1}{3} (x + 2)) + C

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{1}{3} (x + 2))

Simplify

= \frac{1}{3} arcsec (\frac{1}{3} (x + 2))

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{1}{3} (x + 2)) + C

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