What is the integral of (14x-61)/(sqrt(x^2-10x+41)) ?

The integral of (14x-61)/(sqrt(x^2-10x+41)) is 14sqrt(x^2-10x+41)+9ln(1/4 |x-5+sqrt(x^2-10x+41)|)+C

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integral of (14x-61)/(sqrt(x^2-10x+41))

\int \frac{14 x - 61}{x ^{2} - 10 x + 41} d x

14 x^{2} - 10 x + 41 + 9 ln (\frac{1}{4} x - 5 + x^{2} - 10 x + 41) + C

Solution steps

\int \frac{14 x - 61}{x ^{2} - 10 x + 41} d x

Complete the square

x^{2} - 10 x + 41 : (x - 5)^{2} + 16

= \int \frac{14 x - 61}{( x - 5 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{14 u + 9}{u ^{2} + 16} d u

= \int \frac{14 u + 9}{u ^{2} + 16} d u

Expand

\frac{14 u + 9}{u ^{2} + 16} : \frac{14 u}{u ^{2} + 16} + \frac{9}{u ^{2} + 16}

= \int \frac{14 u}{u ^{2} + 16} + \frac{9}{u ^{2} + 16} 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{14 u}{u ^{2} + 16} d u + \int \frac{9}{u ^{2} + 16} d u

\int \frac{14 u}{u ^{2} + 16} d u = 14 u^{2} + 16

\int \frac{9}{u ^{2} + 16} d u = 9 ln (\frac{1}{4} u + 16 + u^{2})

= 14 u^{2} + 16 + 9 ln (\frac{1}{4} u + 16 + u^{2})

Substitute back

u = x - 5

= 14 (x - 5)^{2} + 16 + 9 ln (\frac{1}{4} x - 5 + 16 + (x - 5)^{2})

Simplify

14 (x - 5)^{2} + 16 + 9 ln (\frac{1}{4} x - 5 + 16 + (x - 5)^{2}) : 14 x^{2} - 10 x + 41 + 9 ln (\frac{1}{4} x - 5 + x^{2} - 10 x + 41)

= 14 x^{2} - 10 x + 41 + 9 ln (\frac{1}{4} x - 5 + x^{2} - 10 x + 41)

Add a constant to the solution

= 14 x^{2} - 10 x + 41 + 9 ln (\frac{1}{4} x - 5 + x^{2} - 10 x + 41) + C

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What is the integral of (14x-61)/(sqrt(x^2-10x+41)) ?
The integral of (14x-61)/(sqrt(x^2-10x+41)) is 14sqrt(x^2-10x+41)+9ln(1/4 |x-5+sqrt(x^2-10x+41)|)+C