What is the integral of (10x-17)/(sqrt(x^2-6x-55)) ?

The integral of (10x-17)/(sqrt(x^2-6x-55)) is 10sqrt(x^2-6x-55)+13ln(1/8 |sqrt(x^2-6x-55)+x-3|)+C

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integral of (10x-17)/(sqrt(x^2-6x-55))

\int \frac{10 x - 17}{x ^{2} - 6 x - 55} d x

10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3) + C

Solution steps

\int \frac{10 x - 17}{x ^{2} - 6 x - 55} d x

Complete the square

x^{2} - 6 x - 55 : (x - 3)^{2} - 64

= \int \frac{10 x - 17}{( x - 3 ) ^{2} - 64} d x

Apply u-substitution

: \int \frac{10 u + 13}{u ^{2} - 64} d u

= \int \frac{10 u + 13}{u ^{2} - 64} d u

Expand

\frac{10 u + 13}{u ^{2} - 64} : \frac{10 u}{u ^{2} - 64} + \frac{13}{u ^{2} - 64}

= \int \frac{10 u}{u ^{2} - 64} + \frac{13}{u ^{2} - 64} 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{10 u}{u ^{2} - 64} d u + \int \frac{13}{u ^{2} - 64} d u

\int \frac{10 u}{u ^{2} - 64} d u = 10 u^{2} - 64

\int \frac{13}{u ^{2} - 64} d u = 13 ln (\frac{1}{8} u^{2} - 64 + u)

= 10 u^{2} - 64 + 13 ln (\frac{1}{8} u^{2} - 64 + u)

Substitute back

u = x - 3

= 10 (x - 3)^{2} - 64 + 13 ln (\frac{1}{8} (x - 3)^{2} - 64 + x - 3)

Simplify

10 (x - 3)^{2} - 64 + 13 ln (\frac{1}{8} (x - 3)^{2} - 64 + x - 3) : 10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3)

= 10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3)

Add a constant to the solution

= 10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3) + C

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

6 x^{l n (x)}

\int_{1}^{9} π (x - 1) d x

\frac{d}{d x} ((arccos (x))^{2})

f (x) = \frac{5}{x}, (5, 1)

What is the integral of (10x-17)/(sqrt(x^2-6x-55)) ?
The integral of (10x-17)/(sqrt(x^2-6x-55)) is 10sqrt(x^2-6x-55)+13ln(1/8 |sqrt(x^2-6x-55)+x-3|)+C