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

The integral from 0 to 1 of 1/(x^2+2x+5) is 1/2 (pi/4-arctan(1/2))

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

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

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

Decimal Notation

0.16087 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int_{\frac{1}{2}}^{1} \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{1}{2}}^{1} \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{1}{2}}^{1}

Compute the boundaries:

\frac{π}{4} - arctan (\frac{1}{2})

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

\int_{0}^{1} sin^{2} (π) x d x

\int_{2}^{3} \frac{6}{2 x ^{2} + 2 x - 4} d x

\int_{0}^{1} 3 x^{2} x^{3} - 5 d x

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

\int_{0}^{2} 5 x^{2} - 11 x d x

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