What is the integral from 4 to 6 of 1/(2x^2-16x+40) ?

The integral from 4 to 6 of 1/(2x^2-16x+40) is pi/(16)

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integral from 4 to 6 of 1/(2x^2-16x+40)

\int_{4}^{6} \frac{1}{2 x ^{2} - 16 x + 40} d x

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

\int_{4}^{6} \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_{4}^{6} \frac{1}{2 ( x - 4 ) ^{2} + 8} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{2 u ^{2} + 8} d u

= \int_{0}^{2} \frac{1}{2 u ^{2} + 8} d u

Apply Integral Substitution

= \int_{0}^{1} \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_{0}^{1} \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)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Simplify

= \frac{π}{16}

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What is the integral from 4 to 6 of 1/(2x^2-16x+40) ?
The integral from 4 to 6 of 1/(2x^2-16x+40) is pi/(16)