What is the integral of 7/((x-3)sqrt(x^2-6x+5)) ?

The integral of 7/((x-3)sqrt(x^2-6x+5)) is 7/2 arcsec(1/2 (x-3))+C

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integral of 7/((x-3)sqrt(x^2-6x+5))

\int \frac{7}{( x - 3 ) x ^{2} - 6 x + 5} d x

\frac{7}{2} arcsec (\frac{1}{2} (x - 3)) + C

Solution steps

\int \frac{7}{( x - 3 ) x ^{2} - 6 x + 5} 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 - 3 ) x ^{2} - 6 x + 5} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 4} d u

= 7 \cdot \int \frac{1}{u u ^{2} - 4} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 7 \cdot \int \frac{1}{2} d v

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 7 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= 7 \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (x - 3))

Simplify

7 \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (x - 3)) : \frac{7}{2} arcsec (\frac{1}{2} (x - 3))

= \frac{7}{2} arcsec (\frac{1}{2} (x - 3))

Add a constant to the solution

= \frac{7}{2} arcsec (\frac{1}{2} (x - 3)) + C

\frac{\partial}{\partial x} (e^{- x} sin (x))

\frac{d}{d x} (\frac{8}{x - 5})

f (x) = - csc^{2} (x)

(x^{2} - y^{2}) d x + (2 x y) d y = 0

\int \frac{1}{- x ^{2} + 4 x - 3} d x

What is the integral of 7/((x-3)sqrt(x^2-6x+5)) ?
The integral of 7/((x-3)sqrt(x^2-6x+5)) is 7/2 arcsec(1/2 (x-3))+C