What is the integral of 1/(5x^2-20x+100) ?

The integral of 1/(5x^2-20x+100) is 1/20 arctan((x-2)/4)+C

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integral of 1/(5x^2-20x+100)

\int \frac{1}{5 x ^{2} - 20 x + 100} d x

\frac{1}{20} arctan (\frac{x - 2}{4}) + C

Solution steps

\int \frac{1}{5 x ^{2} - 20 x + 100} d x

Complete the square

5 x^{2} - 20 x + 100 : 5 (x - 2)^{2} + 80

= \int \frac{1}{5 ( x - 2 ) ^{2} + 80} d x

Apply u-substitution

= \int \frac{1}{5 u ^{2} + 80} d u

Apply Integral Substitution

= \int \frac{1}{20 ( 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}{20} \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}{20} arctan (v)

Substitute back

= \frac{1}{20} arctan (\frac{5}{4 5} (x - 2))

Simplify

\frac{1}{20} arctan (\frac{5}{4 5} (x - 2)) : \frac{1}{20} arctan (\frac{x - 2}{4})

= \frac{1}{20} arctan (\frac{x - 2}{4})

Add a constant to the solution

= \frac{1}{20} arctan (\frac{x - 2}{4}) + C

What is the integral of 1/(5x^2-20x+100) ?
The integral of 1/(5x^2-20x+100) is 1/20 arctan((x-2)/4)+C