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

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

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

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

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

Solution steps

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

Complete the square

x^{2} - 2 x + 4 : (x - 1)^{2} + 3

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

Add a constant to the solution

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

f (x) = cos^{6} (x)

\frac{\partial}{\partial x} (\frac{y}{( 1 + x ) ^{2}})

x \to - 6 lim (- 9 x)

e^{5 x}, e^{7 x}, x = 1

y = \frac{2}{x}, (1, 2)

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