What is the integral of ((x-4))/(sqrt(4x^2-32x-1)) ?

The integral of ((x-4))/(sqrt(4x^2-32x-1)) is 1/4 sqrt(4x^2-32x-1)+C

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integral of ((x-4))/(sqrt(4x^2-32x-1))

\int \frac{( x - 4 )}{4 x ^{2} - 32 x - 1} d x

\frac{1}{4} 4 x^{2} - 32 x - 1 + C

Solution steps

\int \frac{x - 4}{4 x ^{2} - 32 x - 1} d x

Apply u-substitution

: \int \frac{u}{4 u ^{2} - 65} d u

= \int \frac{u}{4 u ^{2} - 65} d u

Apply u-substitution

: \int \frac{1}{8 v} d v

= \int \frac{1}{8 v} d v

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= \frac{1}{8} \cdot 2 v^{\frac{1}{2}}

Substitute back

= \frac{1}{8} \cdot 2 (4 (x - 4)^{2} - 65)^{\frac{1}{2}}

Simplify

\frac{1}{8} \cdot 2 (4 (x - 4)^{2} - 65)^{\frac{1}{2}} : \frac{1}{4} 4 x^{2} - 32 x - 1

= \frac{1}{4} 4 x^{2} - 32 x - 1

Add a constant to the solution

= \frac{1}{4} 4 x^{2} - 32 x - 1 + C

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What is the integral of ((x-4))/(sqrt(4x^2-32x-1)) ?
The integral of ((x-4))/(sqrt(4x^2-32x-1)) is 1/4 sqrt(4x^2-32x-1)+C