What is the integral of 1/(4x^2-8x+32) ?

The integral of 1/(4x^2-8x+32) is 1/(4sqrt(7))arctan((x-1)/(sqrt(7)))+C

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integral of 1/(4x^2-8x+32)

\int \frac{1}{4 x ^{2} - 8 x + 32} d x

\frac{1}{4 7} arctan (\frac{x - 1}{7}) + C

Solution steps

\int \frac{1}{4 x ^{2} - 8 x + 32} d x

Complete the square

4 x^{2} - 8 x + 32 : 4 (x - 1)^{2} + 28

= \int \frac{1}{4 ( x - 1 ) ^{2} + 28} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 28} d u

= \int \frac{1}{4 u ^{2} + 28} d u

Apply Integral Substitution

= \int \frac{1}{4 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}{4 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}{4 7} arctan (v)

Substitute back

= \frac{1}{4 7} arctan (\frac{2}{2 7} (x - 1))

Simplify

\frac{1}{4 7} arctan (\frac{2}{2 7} (x - 1)) : \frac{1}{4 7} arctan (\frac{x - 1}{7})

= \frac{1}{4 7} arctan (\frac{x - 1}{7})

Add a constant to the solution

= \frac{1}{4 7} arctan (\frac{x - 1}{7}) + C

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What is the integral of 1/(4x^2-8x+32) ?
The integral of 1/(4x^2-8x+32) is 1/(4sqrt(7))arctan((x-1)/(sqrt(7)))+C