What is the integral of 1/(sqrt(x^2-4x-21)) ?

The integral of 1/(sqrt(x^2-4x-21)) is ln(1/5 |sqrt(x^2-4x-21)+x-2|)+C

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integral of 1/(sqrt(x^2-4x-21))

\int \frac{1}{x ^{2} - 4 x - 21} d x

ln (\frac{1}{5} x^{2} - 4 x - 21 + x - 2) + C

Solution steps

\int \frac{1}{x ^{2} - 4 x - 21} d x

Complete the square

x^{2} - 4 x - 21 : (x - 2)^{2} - 25

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

Simplify

ln tan (arcsec (\frac{1}{5} (x - 2))) + sec (arcsec (\frac{1}{5} (x - 2))) : ln (\frac{1}{5} x^{2} - 4 x - 21 + x - 2)

= ln (\frac{1}{5} x^{2} - 4 x - 21 + x - 2)

Add a constant to the solution

= ln (\frac{1}{5} x^{2} - 4 x - 21 + x - 2) + C

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

\int \frac{5}{e ^{4 x}} d x

\int \frac{x - 76}{x ^{2} ( x + 4 )} d x

\int \frac{tan ^{4} ( x )}{cos ^{4} ( x )} d x

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

What is the integral of 1/(sqrt(x^2-4x-21)) ?
The integral of 1/(sqrt(x^2-4x-21)) is ln(1/5 |sqrt(x^2-4x-21)+x-2|)+C