What is the integral of 7/(((x-1)sqrt(x^2-2x-48))) ?

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

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integral of 7/(((x-1)sqrt(x^2-2x-48)))

\int \frac{7}{( ( x - 1 ) x ^{2} - 2 x - 48 )} d x

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

Solution steps

\int \frac{7}{( 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

= 7 \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

= 7 \cdot \int \frac{1}{( x - 1 ) ( x - 1 ) ^{2} - 49} d x

Apply u-substitution

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

= 7 \cdot \int \frac{1}{u u ^{2} - 49} d u

Apply Trigonometric Substitution

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

= 7 \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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = 3 y + t

\frac{1}{π} sin (π x)

\frac{d}{d x} (\frac{3}{x + 5})

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

\frac{5}{( 3 x ^{2} - x ) ^{4}}

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