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

The integral from 1 to 2 of 1/(x^2+2x+5) is 1/2 (arctan(3/2)-pi/4)

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

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

\frac{1}{2} (arctan (\frac{3}{2}) - \frac{π}{4})

Solution steps

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

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int_{1}^{\frac{3}{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_{1}^{\frac{3}{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)]_{1}^{\frac{3}{2}}

Compute the boundaries:

arctan (\frac{3}{2}) - \frac{π}{4}

= \frac{1}{2} (arctan (\frac{3}{2}) - \frac{π}{4})

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