What is the integral of 1/(x^2-4x+7) ?

The integral of 1/(x^2-4x+7) is 1/(sqrt(3))arctan((x-2)/(sqrt(3)))+C

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integral of 1/(x^2-4x+7)

\int \frac{1}{x ^{2} - 4 x + 7} d x

\frac{1}{3} arctan (\frac{x - 2}{3}) + C

Solution steps

\int \frac{1}{x ^{2} - 4 x + 7} d x

Complete the square

x^{2} - 4 x + 7 : (x - 2)^{2} + 3

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

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

Add a constant to the solution

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

y^{^{'}} - \frac{1}{tan ( x )} y = \frac{1}{tan ( x )}

\int (2 x^{5} - 2 x)^{3} (10 x^{4} - 2) d x

\frac{d}{d x} (\frac{x ^{3}}{e ^{x}})

\frac{\partial}{\partial x} (- 3 y cos (- 9 x))

f (x) = \frac{1}{r + 3 e ^{r}}

What is the integral of 1/(x^2-4x+7) ?
The integral of 1/(x^2-4x+7) is 1/(sqrt(3))arctan((x-2)/(sqrt(3)))+C