What is the integral from 0 to 2 of 1/(x^2+x-6) ?

The integral from 0 to 2 of 1/(x^2+x-6) is diverges

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integral from 0 to 2 of 1/(x^2+x-6)

\int_{0}^{2} \frac{1}{x ^{2} + x - 6} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{x ^{2} + x - 6} d x

Complete the square

x^{2} + x - 6 : (x + \frac{1}{2})^{2} - \frac{25}{4}

= \int_{0}^{2} \frac{1}{( x + \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{4}{4 u ^{2} - 25} d u

= \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{4}{4 u ^{2} - 25} d u

Take the constant out:

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

= 4 \cdot \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{1}{4 u ^{2} - 25} d u

Apply Integral Substitution

= 4 \cdot \int_{\frac{1}{5}}^{1} \frac{1}{10 ( 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}{10} \cdot \int_{\frac{1}{5}}^{1} \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 4 \cdot \frac{1}{10} \cdot \int_{\frac{1}{5}}^{1} \frac{1}{- ( - 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}{10} (- \int_{\frac{1}{5}}^{1} \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 4 \cdot \frac{1}{10} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{5}}^{1})

Simplify

4 \cdot \frac{1}{10} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{5}}^{1}) : - \frac{2}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{1}{5}}^{1}

= - \frac{2}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{1}{5}}^{1}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (e^{- 2 x} cos (3 x))

x \to 11 lim (10)

\int (4 sec^{3} (x)) d x

n = - 4 \sum \infty (\frac{- 4}{9})^{n}

\int 5^{s i n (2 x)} cos (2 x) d x

What is the integral from 0 to 2 of 1/(x^2+x-6) ?
The integral from 0 to 2 of 1/(x^2+x-6) is diverges