What is the integral of (6x+5)/(x^2-20x+101) ?

The integral of (6x+5)/(x^2-20x+101) is 3ln|x^2-20x+101|+65arctan(x-10)+C

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integral of (6x+5)/(x^2-20x+101)

\int \frac{6 x + 5}{x ^{2} - 20 x + 101} d x

3 ln x^{2} - 20 x + 101 + 65 arctan (x - 10) + C

Solution steps

\int \frac{6 x + 5}{x ^{2} - 20 x + 101} d x

Complete the square

x^{2} - 20 x + 101 : (x - 10)^{2} + 1

= \int \frac{6 x + 5}{( x - 10 ) ^{2} + 1} d x

Apply u-substitution

: \int \frac{6 u + 65}{u ^{2} + 1} d u

= \int \frac{6 u + 65}{u ^{2} + 1} d u

Expand

\frac{6 u + 65}{u ^{2} + 1} : \frac{6 u}{u ^{2} + 1} + \frac{65}{u ^{2} + 1}

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{6 u}{u ^{2} + 1} d u + \int \frac{65}{u ^{2} + 1} d u

\int \frac{6 u}{u ^{2} + 1} d u = 3 ln u^{2} + 1

\int \frac{65}{u ^{2} + 1} d u = 65 arctan (u)

= 3 ln u^{2} + 1 + 65 arctan (u)

Substitute back

u = x - 10

= 3 ln (x - 10)^{2} + 1 + 65 arctan (x - 10)

Expand

(x - 10)^{2} + 1 : x^{2} - 20 x + 101

= 3 ln x^{2} - 20 x + 101 + 65 arctan (x - 10)

Add a constant to the solution

= 3 ln x^{2} - 20 x + 101 + 65 arctan (x - 10) + C

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What is the integral of (6x+5)/(x^2-20x+101) ?
The integral of (6x+5)/(x^2-20x+101) is 3ln|x^2-20x+101|+65arctan(x-10)+C