What is the integral of x/(sqrt(x^2-8x-7)) ?

The integral of x/(sqrt(x^2-8x-7)) is sqrt(x^2-8x-7)+4ln|sqrt(1/23 (x^2-8x-7))+1/(sqrt(23))(x-4)|+C

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integral of x/(sqrt(x^2-8x-7))

\int \frac{x}{x ^{2} - 8 x - 7} d x

x^{2} - 8 x - 7 + 4 ln \frac{1}{23} (x^{2} - 8 x - 7) + \frac{1}{23} (x - 4) + C

Solution steps

\int \frac{x}{x ^{2} - 8 x - 7} d x

Complete the square

x^{2} - 8 x - 7 : (x - 4)^{2} - 23

= \int \frac{x}{( x - 4 ) ^{2} - 23} d x

Apply u-substitution

: \int \frac{u + 4}{u ^{2} - 23} d u

= \int \frac{u + 4}{u ^{2} - 23} d u

Expand

\frac{u + 4}{u ^{2} - 23} : \frac{u}{u ^{2} - 23} + \frac{4}{u ^{2} - 23}

= \int \frac{u}{u ^{2} - 23} + \frac{4}{u ^{2} - 23} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{u ^{2} - 23} d u + \int \frac{4}{u ^{2} - 23} d u

\int \frac{u}{u ^{2} - 23} d u = u^{2} - 23

\int \frac{4}{u ^{2} - 23} d u = 4 ln \frac{1}{23} (u^{2} - 23) + \frac{1}{23} u

= u^{2} - 23 + 4 ln \frac{1}{23} (u^{2} - 23) + \frac{1}{23} u

Substitute back

u = x - 4

= (x - 4)^{2} - 23 + 4 ln \frac{1}{23} ((x - 4)^{2} - 23) + \frac{1}{23} (x - 4)

Simplify

(x - 4)^{2} - 23 + 4 ln \frac{1}{23} ((x - 4)^{2} - 23) + \frac{1}{23} (x - 4) : x^{2} - 8 x - 7 + 4 ln \frac{1}{23} (x^{2} - 8 x - 7) + \frac{1}{23} (x - 4)

= x^{2} - 8 x - 7 + 4 ln \frac{1}{23} (x^{2} - 8 x - 7) + \frac{1}{23} (x - 4)

Add a constant to the solution

= x^{2} - 8 x - 7 + 4 ln \frac{1}{23} (x^{2} - 8 x - 7) + \frac{1}{23} (x - 4) + C

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What is the integral of x/(sqrt(x^2-8x-7)) ?
The integral of x/(sqrt(x^2-8x-7)) is sqrt(x^2-8x-7)+4ln|sqrt(1/23 (x^2-8x-7))+1/(sqrt(23))(x-4)|+C