What is the integral of 1/(t^2+2t-10) ?

The integral of 1/(t^2+2t-10) is -1/(2sqrt(11))(ln|(t+1)/(sqrt(11))+1|-ln|(t+1)/(sqrt(11))-1|)+C

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integral of 1/(t^2+2t-10)

\int \frac{1}{t ^{2} + 2 t - 10} d t

- \frac{1}{2 11} (ln \frac{t + 1}{11} + 1 - ln \frac{t + 1}{11} - 1) + C

Solution steps

\int \frac{1}{t ^{2} + 2 t - 10} d t

Factor

t^{2} + 2 t - 10 : - (- t^{2} - 2 t + 10)

= \int \frac{1}{- ( - t ^{2} - 2 t + 10 )} d t

Take the constant out:

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

= - \int \frac{1}{- t ^{2} - 2 t + 10} d t

Complete the square

- t^{2} - 2 t + 10 : - (t + 1)^{2} + 11

= - \int \frac{1}{- ( t + 1 ) ^{2} + 11} d t

Apply u-substitution

: \int \frac{1}{- u ^{2} + 11} d u

= - \int \frac{1}{- u ^{2} + 11} d u

Apply Integral Substitution

= - \int \frac{1}{11 ( - 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}{11} \cdot \int \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}

= - \frac{1}{11} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{11} \frac{ln \frac{t + 1}{11} + 1}{2} - \frac{ln \frac{t + 1}{11} - 1}{2}

Simplify

- \frac{1}{11} \frac{ln \frac{t + 1}{11} + 1}{2} - \frac{ln \frac{t + 1}{11} - 1}{2} : - \frac{1}{2 11} (ln \frac{t + 1}{11} + 1 - ln \frac{t + 1}{11} - 1)

= - \frac{1}{2 11} (ln \frac{t + 1}{11} + 1 - ln \frac{t + 1}{11} - 1)

Add a constant to the solution

= - \frac{1}{2 11} (ln \frac{t + 1}{11} + 1 - ln \frac{t + 1}{11} - 1) + C

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What is the integral of 1/(t^2+2t-10) ?
The integral of 1/(t^2+2t-10) is -1/(2sqrt(11))(ln|(t+1)/(sqrt(11))+1|-ln|(t+1)/(sqrt(11))-1|)+C