What is the integral of 1/(x^2-4x+8) ?

The integral of 1/(x^2-4x+8) is 1/2 arctan((x-2)/2)+C

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integral of 1/(x^2-4x+8)

\int \frac{1}{x ^{2} - 4 x + 8} d x

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

Solution steps

\int \frac{1}{x ^{2} - 4 x + 8} d x

Complete the square

x^{2} - 4 x + 8 : (x - 2)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

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

f (x) = (x)^{3}

x \to 0 lim (\frac{( x + 1 ) ^{2} - 1}{x})

y^{^{'}} = 6 x + y, y (0) = 8

\int_{0}^{l n (2)} e^{- x} d x

\frac{d y}{d x} = \frac{sec ( y )}{( x - 1 ) ^{2}}

What is the integral of 1/(x^2-4x+8) ?
The integral of 1/(x^2-4x+8) is 1/2 arctan((x-2)/2)+C