What is the integral of (20)/((x-1)sqrt(x^2-2x-48)) ?

The integral of (20)/((x-1)sqrt(x^2-2x-48)) is 20/7 arcsec(1/7 (x-1))+C

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integral of (20)/((x-1)sqrt(x^2-2x-48))

\int \frac{20}{( x - 1 ) x ^{2} - 2 x - 48} d x

\frac{20}{7} arcsec (\frac{1}{7} (x - 1)) + C

Solution steps

\int \frac{20}{( x - 1 ) x ^{2} - 2 x - 48} d x

Take the constant out:

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

= 20 \cdot \int \frac{1}{( x - 1 ) x ^{2} - 2 x - 48} d x

Complete the square

x^{2} - 2 x - 48 : (x - 1)^{2} - 49

= 20 \cdot \int \frac{1}{( x - 1 ) ( x - 1 ) ^{2} - 49} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 49} d u

= 20 \cdot \int \frac{1}{u u ^{2} - 49} d u

Apply Trigonometric Substitution

: \int \frac{1}{7} d v

= 20 \cdot \int \frac{1}{7} d v

Take the constant out:

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

= 20 \cdot \frac{1}{7} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 20 \cdot \frac{1}{7} \cdot 1 \cdot v

Substitute back

= 20 \cdot \frac{1}{7} \cdot 1 \cdot arcsec (\frac{1}{7} (x - 1))

Simplify

20 \cdot \frac{1}{7} \cdot 1 \cdot arcsec (\frac{1}{7} (x - 1)) : \frac{20}{7} arcsec (\frac{1}{7} (x - 1))

= \frac{20}{7} arcsec (\frac{1}{7} (x - 1))

Add a constant to the solution

= \frac{20}{7} arcsec (\frac{1}{7} (x - 1)) + C

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What is the integral of (20)/((x-1)sqrt(x^2-2x-48)) ?
The integral of (20)/((x-1)sqrt(x^2-2x-48)) is 20/7 arcsec(1/7 (x-1))+C