What is the integral of 1/(2x^2-16x+40) ?

The integral of 1/(2x^2-16x+40) is 1/4 arctan((x-4)/2)+C

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integral of 1/(2x^2-16x+40)

\int \frac{1}{2 x ^{2} - 16 x + 40} d x

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

Solution steps

\int \frac{1}{2 x ^{2} - 16 x + 40} d x

Complete the square

2 x^{2} - 16 x + 40 : 2 (x - 4)^{2} + 8

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

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 8} d u

= \int \frac{1}{2 u ^{2} + 8} d u

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\int 5 (400 + 5 t)^{3} d t

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\int \frac{1}{ln ( 3 x )} d x

What is the integral of 1/(2x^2-16x+40) ?
The integral of 1/(2x^2-16x+40) is 1/4 arctan((x-4)/2)+C