What is the integral of 3/(2x^2-4x+10) ?

The integral of 3/(2x^2-4x+10) is 3/4 arctan((x-1)/2)+C

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integral of 3/(2x^2-4x+10)

\int \frac{3}{2 x ^{2} - 4 x + 10} d x

\frac{3}{4} arctan (\frac{x - 1}{2}) + C

Solution steps

\int \frac{3}{2 x ^{2} - 4 x + 10} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{2 x ^{2} - 4 x + 10} d x

Complete the square

2 x^{2} - 4 x + 10 : 2 (x - 1)^{2} + 8

= 3 \cdot \int \frac{1}{2 ( x - 1 ) ^{2} + 8} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{2 u ^{2} + 8} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{4 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 3 \cdot \frac{1}{4} arctan (v)

Substitute back

= 3 \cdot \frac{1}{4} arctan (\frac{2}{2 2} (x - 1))

Simplify

3 \cdot \frac{1}{4} arctan (\frac{2}{2 2} (x - 1)) : \frac{3}{4} arctan (\frac{x - 1}{2})

= \frac{3}{4} arctan (\frac{x - 1}{2})

Add a constant to the solution

= \frac{3}{4} arctan (\frac{x - 1}{2}) + C

What is the integral of 3/(2x^2-4x+10) ?
The integral of 3/(2x^2-4x+10) is 3/4 arctan((x-1)/2)+C