What is the integral of 5/(x^2+6x+10) ?

The integral of 5/(x^2+6x+10) is 5arctan(x+3)+C

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integral of 5/(x^2+6x+10)

\int \frac{5}{x ^{2} + 6 x + 10} d x

5 arctan (x + 3) + C

Solution steps

\int \frac{5}{x ^{2} + 6 x + 10} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} + 6 x + 10} d x

Complete the square

x^{2} + 6 x + 10 : (x + 3)^{2} + 1

= 5 \cdot \int \frac{1}{( x + 3 ) ^{2} + 1} d x

Apply u-substitution

= 5 \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 5 arctan (u)

Substitute back

u = x + 3

= 5 arctan (x + 3)

Add a constant to the solution

= 5 arctan (x + 3) + C

What is the integral of 5/(x^2+6x+10) ?
The integral of 5/(x^2+6x+10) is 5arctan(x+3)+C