What is the integral of 4/(x^2-4x+9) ?

The integral of 4/(x^2-4x+9) is 4/(sqrt(5))arctan(1/(sqrt(5))(x-2))+C

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integral of 4/(x^2-4x+9)

\int \frac{4}{x ^{2} - 4 x + 9} d x

\frac{4}{5} arctan (\frac{1}{5} (x - 2)) + C

Solution steps

\int \frac{4}{x ^{2} - 4 x + 9} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} - 4 x + 9} d x

Complete the square

x^{2} - 4 x + 9 : (x - 2)^{2} + 5

= 4 \cdot \int \frac{1}{( x - 2 ) ^{2} + 5} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 5} d u

= 4 \cdot \int \frac{1}{u ^{2} + 5} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{5} arctan (v)

Substitute back

= 4 \cdot \frac{1}{5} arctan (\frac{x - 2}{5})

Simplify

4 \cdot \frac{1}{5} arctan (\frac{x - 2}{5}) : \frac{4}{5} arctan (\frac{1}{5} (x - 2))

= \frac{4}{5} arctan (\frac{1}{5} (x - 2))

Add a constant to the solution

= \frac{4}{5} arctan (\frac{1}{5} (x - 2)) + C

(\frac{x ^{'}}{x})^{^{'}}

\int \frac{72}{w ^{2} 36 - w ^{2}} d w

y^{^{'}} + sin (t) y = sin (t), y (0) = 8

\frac{\partial}{\partial x} (3 x^{2} y - 4 x y^{2})

\frac{t}{( t - 4 ) ^{2}}

What is the integral of 4/(x^2-4x+9) ?
The integral of 4/(x^2-4x+9) is 4/(sqrt(5))arctan(1/(sqrt(5))(x-2))+C