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

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

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

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

arctan (x + 3) + C

Solution steps

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

Complete the square

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

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

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 = x + 3

= arctan (x + 3)

Add a constant to the solution

= arctan (x + 3) + C

(10 + x^{6}) \frac{d y}{d x} = \frac{x ^{5}}{y}

x \to 2 lim (x (x + 1) (x + 2))

x \frac{d y}{d x} - y = 0

\frac{1}{s ^{2} ( s ^{2} + 5 )}

\int cos (1 - 2 x) d x

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