What is the integral of 1/(x-8sqrt(x)+15) ?

The integral of 1/(x-8sqrt(x)+15) is ln|x-8sqrt(x)+15|-4ln|sqrt(x)-3|+4ln|sqrt(x)-5|+C

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integral of 1/(x-8sqrt(x)+15)

\int \frac{1}{x - 8 x + 15} d x

ln x - 8 x + 15 - 4 ln x - 3 + 4 ln x - 5 + C

Solution steps

\int \frac{1}{x - 8 x + 15} d x

Apply u-substitution

= \int \frac{2 u}{u ^{2} - 8 u + 15} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u}{u ^{2} - 8 u + 15} d u

Complete the square

u^{2} - 8 u + 15 : (u - 4)^{2} - 1

= 2 \cdot \int \frac{u}{( u - 4 ) ^{2} - 1} d u

Apply u-substitution

= 2 \cdot \int \frac{v + 4}{v ^{2} - 1} d v

Expand

\frac{v + 4}{v ^{2} - 1} : \frac{v}{v ^{2} - 1} + \frac{4}{v ^{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

= 2 (\int \frac{v}{v ^{2} - 1} d v + \int \frac{4}{v ^{2} - 1} d v)

\int \frac{v}{v ^{2} - 1} d v = \frac{1}{2} ln v^{2} - 1

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

= 2 (\frac{1}{2} ln v^{2} - 1 - 4 (\frac{1}{2} ln ∣ v + 1 ∣ - \frac{1}{2} ln ∣ v - 1 ∣))

Substitute back

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

Simplify

2 (\frac{1}{2} ln (x - 4)^{2} - 1 - 4 (\frac{1}{2} ln x - 4 + 1 - \frac{1}{2} ln x - 4 - 1)) : ln x - 8 x + 15 - 4 ln x - 3 + 4 ln x - 5

= ln x - 8 x + 15 - 4 ln x - 3 + 4 ln x - 5

Add a constant to the solution

= ln x - 8 x + 15 - 4 ln x - 3 + 4 ln x - 5 + C

What is the integral of 1/(x-8sqrt(x)+15) ?
The integral of 1/(x-8sqrt(x)+15) is ln|x-8sqrt(x)+15|-4ln|sqrt(x)-3|+4ln|sqrt(x)-5|+C