What is the integral of 3/(2x^2+8x+10) ?

The integral of 3/(2x^2+8x+10) is 3/2 arctan(x+2)+C

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integral of 3/(2x^2+8x+10)

\int \frac{3}{2 x ^{2} + 8 x + 10} d x

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

Solution steps

\int \frac{3}{2 x ^{2} + 8 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} + 8 x + 10} d x

Complete the square

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

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

Apply u-substitution

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

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

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

= 3 \cdot \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

= 3 \cdot \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)

= 3 \cdot \frac{1}{2} arctan (u)

Substitute back

u = x + 2

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

Simplify

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

= \frac{3}{2} arctan (x + 2)

Add a constant to the solution

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

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What is the integral of 3/(2x^2+8x+10) ?
The integral of 3/(2x^2+8x+10) is 3/2 arctan(x+2)+C