What is the integral of 1/(4x^2-16x+17) ?

The integral of 1/(4x^2-16x+17) is 1/2 arctan(2(x-2))+C

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integral of 1/(4x^2-16x+17)

\int \frac{1}{4 x ^{2} - 16 x + 17} d x

\frac{1}{2} arctan (2 (x - 2)) + C

Solution steps

\int \frac{1}{4 x ^{2} - 16 x + 17} d x

Complete the square

4 x^{2} - 16 x + 17 : 4 (x - 2)^{2} + 1

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

Apply u-substitution

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

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

= \frac{1}{2} arctan (2 (x - 2)) + C

What is the integral of 1/(4x^2-16x+17) ?
The integral of 1/(4x^2-16x+17) is 1/2 arctan(2(x-2))+C