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

\frac{d}{d x} (x^{3} + (3 - 3 x^{2})^{2} + 4)

\frac{d y}{d x} + y = x^{2} - 2

x \to 2 lim (sin (\frac{π x}{2}))

\int \frac{1}{u ^{\frac{3}{2}}} d u

2 sin^{2} (6 x^{4})

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