What is the integral of 1/(y^2+6y+10) ?

The integral of 1/(y^2+6y+10) is arctan(y+3)+C

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integral of 1/(y^2+6y+10)

\int \frac{1}{y ^{2} + 6 y + 10} d y

arctan (y + 3) + C

Solution steps

\int \frac{1}{y ^{2} + 6 y + 10} d y

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = y + 3

= arctan (y + 3)

Add a constant to the solution

= arctan (y + 3) + C

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What is the integral of 1/(y^2+6y+10) ?
The integral of 1/(y^2+6y+10) is arctan(y+3)+C