What is the integral of (x-4)/(sqrt(x^2-8x+1)) ?

The integral of (x-4)/(sqrt(x^2-8x+1)) is sqrt(x^2-8x+1)+C

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integral of (x-4)/(sqrt(x^2-8x+1))

\int \frac{x - 4}{x ^{2} - 8 x + 1} d x

x^{2} - 8 x + 1 + C

Solution steps

\int \frac{x - 4}{x ^{2} - 8 x + 1} d x

Apply u-substitution

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

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} - 8 x + 1

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

Simplify

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

= x^{2} - 8 x + 1

Add a constant to the solution

= x^{2} - 8 x + 1 + C

t \to 3 - lim (\frac{t ^{2}}{9 - t ^{2}})

\int \frac{x ^{2}}{x ^{2} + 8} d x

\frac{e ^{x} - 1}{x ^{2} + 1}

y^{^{''}} - 4 y^{^{'}} + 4 = 0

n = 0 \sum \infty \frac{( 3 ^{n} )}{n !}

What is the integral of (x-4)/(sqrt(x^2-8x+1)) ?
The integral of (x-4)/(sqrt(x^2-8x+1)) is sqrt(x^2-8x+1)+C