What is the integral from-2 to 2 of 1/(x^2+4x+13) ?

The integral from-2 to 2 of 1/(x^2+4x+13) is 1/3 arctan(4/3)

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integral from-2 to 2 of 1/(x^2+4x+13)

\int_{- 2}^{2} \frac{1}{x ^{2} + 4 x + 13} d x

\frac{1}{3} arctan (\frac{4}{3})

Decimal Notation

0.30909 \dots

Solution steps

\int_{- 2}^{2} \frac{1}{x ^{2} + 4 x + 13} d x

Complete the square

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= \int_{- 2}^{2} \frac{1}{( x + 2 ) ^{2} + 9} d x

Apply u-substitution

: \int_{0}^{4} \frac{1}{u ^{2} + 9} d u

= \int_{0}^{4} \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= \int_{0}^{\frac{4}{3}} \frac{1}{3 ( 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}{3} \cdot \int_{0}^{\frac{4}{3}} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{3} [arctan (v)]_{0}^{\frac{4}{3}}

Compute the boundaries:

arctan (\frac{4}{3})

= \frac{1}{3} arctan (\frac{4}{3})

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\frac{d}{d x} (8 sin^{2} (x))

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\int (4 + x^{7})^{3} 7 x^{6} d x

What is the integral from-2 to 2 of 1/(x^2+4x+13) ?
The integral from-2 to 2 of 1/(x^2+4x+13) is 1/3 arctan(4/3)