What is the integral of 4/(sqrt(x^2-8x-9)) ?

The integral of 4/(sqrt(x^2-8x-9)) is 4ln(1/5 |sqrt(x^2-8x-9)+x-4|)+C

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integral of 4/(sqrt(x^2-8x-9))

\int \frac{4}{x ^{2} - 8 x - 9} d x

4 ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4) + C

Solution steps

\int \frac{4}{x ^{2} - 8 x - 9} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} - 8 x - 9} d x

Complete the square

x^{2} - 8 x - 9 : (x - 4)^{2} - 25

= 4 \cdot \int \frac{1}{( x - 4 ) ^{2} - 25} d x

Apply u-substitution

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

= 4 \cdot \int \frac{1}{u ^{2} - 25} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= 4 \cdot \int sec (v) d v

Use the common integral:

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

= 4 ln ∣ tan (v) + sec (v) ∣

Substitute back

= 4 ln tan (arcsec (\frac{1}{5} (x - 4))) + sec (arcsec (\frac{1}{5} (x - 4)))

Simplify

4 ln tan (arcsec (\frac{1}{5} (x - 4))) + sec (arcsec (\frac{1}{5} (x - 4))) : 4 ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4)

= 4 ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4)

Add a constant to the solution

= 4 ln (\frac{1}{5} x^{2} - 8 x - 9 + x - 4) + C

\frac{d}{d x} (\frac{11 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}})

\frac{x}{x + 1}

\frac{d y}{d x} sin (x) = y ln (y)

x \to \infty lim (- \frac{1}{4})

x \to - 2 + lim (x - 2)

What is the integral of 4/(sqrt(x^2-8x-9)) ?
The integral of 4/(sqrt(x^2-8x-9)) is 4ln(1/5 |sqrt(x^2-8x-9)+x-4|)+C