What is the integral of (4x-16)/(sqrt(x^2-8x+5)) ?

The integral of (4x-16)/(sqrt(x^2-8x+5)) is 4sqrt(x^2-8x+5)+C

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integral of (4x-16)/(sqrt(x^2-8x+5))

\int \frac{4 x - 16}{x ^{2} - 8 x + 5} d x

4 x^{2} - 8 x + 5 + C

Solution steps

\int \frac{4 x - 16}{x ^{2} - 8 x + 5} d x

Complete the square

x^{2} - 8 x + 5 : (x - 4)^{2} - 11

= \int \frac{4 x - 16}{( x - 4 ) ^{2} - 11} d x

Apply u-substitution

: \int \frac{4 u}{u ^{2} - 11} d u

= \int \frac{4 u}{u ^{2} - 11} d u

Take the constant out:

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

= 4 \cdot \int \frac{u}{u ^{2} - 11} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= 4 \cdot \frac{1}{2} \cdot 2 v^{\frac{1}{2}}

Substitute back

= 4 \cdot \frac{1}{2} \cdot 2 ((x - 4)^{2} - 11)^{\frac{1}{2}}

Simplify

4 \cdot \frac{1}{2} \cdot 2 ((x - 4)^{2} - 11)^{\frac{1}{2}} : 4 x^{2} - 8 x + 5

= 4 x^{2} - 8 x + 5

Add a constant to the solution

= 4 x^{2} - 8 x + 5 + C

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What is the integral of (4x-16)/(sqrt(x^2-8x+5)) ?
The integral of (4x-16)/(sqrt(x^2-8x+5)) is 4sqrt(x^2-8x+5)+C