What is the integral of 5/(x^2-4x+8) ?

The integral of 5/(x^2-4x+8) is 5/2 arctan(1/2 (x-2))+C

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integral of 5/(x^2-4x+8)

\int \frac{5}{x ^{2} - 4 x + 8} d x

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

Solution steps

\int \frac{5}{x ^{2} - 4 x + 8} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} - 4 x + 8} d x

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= 5 \cdot \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

= 5 \cdot \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)

= 5 \cdot \frac{1}{2} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

8^{t}

x \to 2 lim (9)

(1, 0.5), (3, 0)

\int x^{2} - 64 d x

\frac{d}{d a} (\frac{π ( 9 a ^{2} + 6 a + 1 )}{54 a})

What is the integral of 5/(x^2-4x+8) ?
The integral of 5/(x^2-4x+8) is 5/2 arctan(1/2 (x-2))+C