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

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

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integral from 0 to 3 of 1/(x^2-x+5/2)

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

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

Solution steps

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

Simplify

\frac{1}{x ^{2} - x + \frac{5}{2}} : \frac{2}{2 x ^{2} - 2 x + 5}

= \int_{0}^{3} \frac{2}{2 x ^{2} - 2 x + 5} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{3} \frac{1}{2 x ^{2} - 2 x + 5} d x

Complete the square

2 x^{2} - 2 x + 5 : 2 (x - \frac{1}{2})^{2} + \frac{9}{2}

= 2 \cdot \int_{0}^{3} \frac{1}{2 ( x - \frac{1}{2} ) ^{2} + \frac{9}{2}} d x

Apply u-substitution

: \int_{- \frac{1}{2}}^{\frac{5}{2}} \frac{2}{4 u ^{2} + 9} d u

= 2 \cdot \int_{- \frac{1}{2}}^{\frac{5}{2}} \frac{2}{4 u ^{2} + 9} d u

Take the constant out:

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

= 2 \cdot 2 \cdot \int_{- \frac{1}{2}}^{\frac{5}{2}} \frac{1}{4 u ^{2} + 9} d u

Apply Integral Substitution

= 2 \cdot 2 \cdot \int_{- \frac{1}{3}}^{\frac{5}{3}} \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot 2 \cdot \frac{1}{6} \cdot \int_{- \frac{1}{3}}^{\frac{5}{3}} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot 2 \cdot \frac{1}{6} [arctan (v)]_{- \frac{1}{3}}^{\frac{5}{3}}

Simplify

2 \cdot 2 \cdot \frac{1}{6} [arctan (v)]_{- \frac{1}{3}}^{\frac{5}{3}} : \frac{2}{3} [arctan (v)]_{- \frac{1}{3}}^{\frac{5}{3}}

= \frac{2}{3} [arctan (v)]_{- \frac{1}{3}}^{\frac{5}{3}}

Compute the boundaries:

arctan (\frac{5}{3}) + arctan (\frac{1}{3})

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

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