What is the integral from 0 to 5 of 1/(x^2+6x+11) ?

The integral from 0 to 5 of 1/(x^2+6x+11) is (arctan(4sqrt(2))-arctan(3/(sqrt(2))))/(sqrt(2))

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integral from 0 to 5 of 1/(x^2+6x+11)

\int_{0}^{5} \frac{1}{x ^{2} + 6 x + 11} d x

\frac{arctan ( 4 2 ) - arctan ( \frac{3}{2} )}{2}

Decimal

0.18776 \dots

Solution steps

\int_{0}^{5} \frac{1}{x ^{2} + 6 x + 11} d x

Complete the square

x^{2} + 6 x + 11 : (x + 3)^{2} + 2

= \int_{0}^{5} \frac{1}{( x + 3 ) ^{2} + 2} d x

Apply u-substitution

: \int_{3}^{8} \frac{1}{u ^{2} + 2} d u

= \int_{3}^{8} \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

= \int_{\frac{3}{2}}^{42} \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

= \frac{1}{2} \cdot \int_{\frac{3}{2}}^{42} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{2} [arctan (v)]_{\frac{3}{2}}^{42}

Compute the boundaries:

arctan (42) - arctan (\frac{3}{2})

= \frac{1}{2} (arctan (42) - arctan (\frac{3}{2}))

Simplify

= \frac{arctan ( 4 2 ) - arctan ( \frac{3}{2} )}{2}

\int_{0}^{4} \frac{1}{6 t + 1} d t

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What is the integral from 0 to 5 of 1/(x^2+6x+11) ?
The integral from 0 to 5 of 1/(x^2+6x+11) is (arctan(4sqrt(2))-arctan(3/(sqrt(2))))/(sqrt(2))