What is the integral from 0 to 2 of 1/(x^2+6x+13) ?

The integral from 0 to 2 of 1/(x^2+6x+13) is 1/2 (arctan(5/2)-arctan(3/2))

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integral from 0 to 2 of 1/(x^2+6x+13)

\int_{0}^{2} \frac{1}{x ^{2} + 6 x + 13} d x

\frac{1}{2} (arctan (\frac{5}{2}) - arctan (\frac{3}{2}))

Decimal

0.10374 \dots

Solution steps

\int_{0}^{2} \frac{1}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int_{0}^{2} \frac{1}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{3}^{5} \frac{1}{u ^{2} + 4} d u

= \int_{3}^{5} \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= \int_{\frac{3}{2}}^{\frac{5}{2}} \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}}^{\frac{5}{2}} \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}}^{\frac{5}{2}}

Compute the boundaries:

arctan (\frac{5}{2}) - arctan (\frac{3}{2})

= \frac{1}{2} (arctan (\frac{5}{2}) - arctan (\frac{3}{2}))

\int_{- 1}^{2} (- 2 x^{2} + 2 x + 4) d x

\int_{0}^{\frac{π}{5}} t^{2} sin (5 t) d t

\int_{1}^{2} (2 x + 1)^{2} d x

\int_{0}^{2} (- 2 x^{2} + 4 x) d x

\int_{- 2}^{2} x (x^{2} - 6) d x

What is the integral from 0 to 2 of 1/(x^2+6x+13) ?
The integral from 0 to 2 of 1/(x^2+6x+13) is 1/2 (arctan(5/2)-arctan(3/2))