What is the integral of 1/(8x^2-16x+64) ?

The integral of 1/(8x^2-16x+64) is 1/(8sqrt(7))arctan((x-1)/(sqrt(7)))+C

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integral of 1/(8x^2-16x+64)

\int \frac{1}{8 x ^{2} - 16 x + 64} d x

\frac{1}{8 7} arctan (\frac{x - 1}{7}) + C

Solution steps

\int \frac{1}{8 x ^{2} - 16 x + 64} d x

Complete the square

8 x^{2} - 16 x + 64 : 8 (x - 1)^{2} + 56

= \int \frac{1}{8 ( x - 1 ) ^{2} + 56} d x

Apply u-substitution

: \int \frac{1}{8 u ^{2} + 56} d u

= \int \frac{1}{8 u ^{2} + 56} d u

Apply Integral Substitution

= \int \frac{1}{8 7 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{8 7} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{8 7} arctan (v)

Substitute back

= \frac{1}{8 7} arctan (\frac{2 2}{2 14} (x - 1))

Simplify

\frac{1}{8 7} arctan (\frac{2 2}{2 14} (x - 1)) : \frac{1}{8 7} arctan (\frac{x - 1}{7})

= \frac{1}{8 7} arctan (\frac{x - 1}{7})

Add a constant to the solution

= \frac{1}{8 7} arctan (\frac{x - 1}{7}) + C

\frac{\partial}{\partial t} (s e^{t})

\frac{\partial}{\partial t} (s sin (t))

x \to - 2 lim (\frac{5}{x + 2})

f (x) = 2^{3 x^{2} - 6}

f (x) = - 6 x^{2} - 2 x, at x = 3

What is the integral of 1/(8x^2-16x+64) ?
The integral of 1/(8x^2-16x+64) is 1/(8sqrt(7))arctan((x-1)/(sqrt(7)))+C