What is the integral of 1/(sqrt(x^2-4x-3)) ?

The integral of 1/(sqrt(x^2-4x-3)) is ln|sqrt(1/7 (x^2-4x-3))+1/(sqrt(7))(x-2)|+C

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integral of 1/(sqrt(x^2-4x-3))

\int \frac{1}{x ^{2} - 4 x - 3} d x

ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2) + C

Solution steps

\int \frac{1}{x ^{2} - 4 x - 3} d x

Complete the square

x^{2} - 4 x - 3 : (x - 2)^{2} - 7

= \int \frac{1}{( x - 2 ) ^{2} - 7} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 7} d u

= \int \frac{1}{u ^{2} - 7} 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}{7} (x - 2))) + sec (arcsec (\frac{1}{7} (x - 2)))

Simplify

ln tan (arcsec (\frac{1}{7} (x - 2))) + sec (arcsec (\frac{1}{7} (x - 2))) : ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2)

= ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2)

Add a constant to the solution

= ln \frac{1}{7} (x^{2} - 4 x - 3) + \frac{1}{7} (x - 2) + C

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cos^{2} (t)

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\int sin (a x) e^{b x + c} d x

What is the integral of 1/(sqrt(x^2-4x-3)) ?
The integral of 1/(sqrt(x^2-4x-3)) is ln|sqrt(1/7 (x^2-4x-3))+1/(sqrt(7))(x-2)|+C