What is the integral from 0 to 2 of 1/(x^2+3x+4) ?

The integral from 0 to 2 of 1/(x^2+3x+4) is (2(arctan(sqrt(7))-arctan(3/(sqrt(7)))))/(sqrt(7))

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integral from 0 to 2 of 1/(x^2+3x+4)

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

\frac{2 ( arctan ( 7 ) - arctan ( \frac{3}{7} ) )}{7}

Decimal Notation

0.27316 \dots

Solution steps

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

Complete the square

x^{2} + 3 x + 4 : (x + \frac{3}{2})^{2} + \frac{7}{4}

= \int_{0}^{2} \frac{1}{( x + \frac{3}{2} ) ^{2} + \frac{7}{4}} d x

Apply u-substitution

: \int_{\frac{3}{2}}^{\frac{7}{2}} \frac{4}{4 u ^{2} + 7} d u

= \int_{\frac{3}{2}}^{\frac{7}{2}} \frac{4}{4 u ^{2} + 7} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{\frac{3}{2}}^{\frac{7}{2}} \frac{1}{4 u ^{2} + 7} d u

Apply Integral Substitution

= 4 \cdot \int_{\frac{3}{7}}^{7} \frac{1}{2 7 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2 7} \cdot \int_{\frac{3}{7}}^{7} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{2 7} [arctan (v)]_{\frac{3}{7}}^{7}

Simplify

4 \cdot \frac{1}{2 7} [arctan (v)]_{\frac{3}{7}}^{7} : \frac{2}{7} [arctan (v)]_{\frac{3}{7}}^{7}

= \frac{2}{7} [arctan (v)]_{\frac{3}{7}}^{7}

Compute the boundaries:

arctan (7) - arctan (\frac{3}{7})

= \frac{2}{7} (arctan (7) - arctan (\frac{3}{7}))

Simplify

= \frac{2 ( arctan ( 7 ) - arctan ( \frac{3}{7} ) )}{7}

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What is the integral from 0 to 2 of 1/(x^2+3x+4) ?
The integral from 0 to 2 of 1/(x^2+3x+4) is (2(arctan(sqrt(7))-arctan(3/(sqrt(7)))))/(sqrt(7))