What is the integral of 1/(sqrt(x^2+2x-24)) ?

The integral of 1/(sqrt(x^2+2x-24)) is ln(1/5 |sqrt(x^2+2x-24)+x+1|)+C

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integral of 1/(sqrt(x^2+2x-24))

\int \frac{1}{x ^{2} + 2 x - 24} d x

ln (\frac{1}{5} x^{2} + 2 x - 24 + x + 1) + C

Solution steps

\int \frac{1}{x ^{2} + 2 x - 24} d x

Complete the square

x^{2} + 2 x - 24 : (x + 1)^{2} - 25

= \int \frac{1}{( x + 1 ) ^{2} - 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 25} d u

= \int \frac{1}{u ^{2} - 25} 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}{5} (x + 1))) + sec (arcsec (\frac{1}{5} (x + 1)))

Simplify

ln tan (arcsec (\frac{1}{5} (x + 1))) + sec (arcsec (\frac{1}{5} (x + 1))) : ln (\frac{1}{5} x^{2} + 2 x - 24 + x + 1)

= ln (\frac{1}{5} x^{2} + 2 x - 24 + x + 1)

Add a constant to the solution

= ln (\frac{1}{5} x^{2} + 2 x - 24 + x + 1) + C

\frac{\partial}{\partial y} (\frac{( x ^{2} y )}{x ^{2} + y ^{2}})

6 x^{2} + 2 x y + 2 y^{2} = 74, (3, 2)

\int \frac{1}{1 - 36 x ^{2}} d x

y^{^{'''}} - 4 y^{^{''}} - y^{^{'}} + 4 y = 0

What is the integral of 1/(sqrt(x^2+2x-24)) ?
The integral of 1/(sqrt(x^2+2x-24)) is ln(1/5 |sqrt(x^2+2x-24)+x+1|)+C