What is the integral of 1/(2x^2-8x+10) ?

The integral of 1/(2x^2-8x+10) is 1/2 arctan(x-2)+C

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integral of 1/(2x^2-8x+10)

\int \frac{1}{2 x ^{2} - 8 x + 10} d x

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

Solution steps

\int \frac{1}{2 x ^{2} - 8 x + 10} d x

Complete the square

2 x^{2} - 8 x + 10 : 2 (x - 2)^{2} + 2

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

Apply u-substitution

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

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

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

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

Use the common integral:

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

= \frac{1}{2} arctan (u)

Substitute back

u = x - 2

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

Add a constant to the solution

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

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What is the integral of 1/(2x^2-8x+10) ?
The integral of 1/(2x^2-8x+10) is 1/2 arctan(x-2)+C