What is the integral of 1/(15+4t-t^2) ?

The integral of 1/(15+4t-t^2) is 1/(2sqrt(19))(ln|(t-2)/(sqrt(19))+1|-ln|(t-2)/(sqrt(19))-1|)+C

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integral of 1/(15+4t-t^2)

\int \frac{1}{15 + 4 t - t ^{2}} d t

\frac{1}{2 19} (ln \frac{t - 2}{19} + 1 - ln \frac{t - 2}{19} - 1) + C

Solution steps

\int \frac{1}{15 + 4 t - t ^{2}} d t

Complete the square

15 + 4 t - t^{2} : - (t - 2)^{2} + 19

= \int \frac{1}{- ( t - 2 ) ^{2} + 19} d t

Apply u-substitution

: \int \frac{1}{- u ^{2} + 19} d u

= \int \frac{1}{- u ^{2} + 19} d u

Apply Integral Substitution

= \int \frac{1}{19 ( - 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}{19} \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}{19} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{19} \frac{ln \frac{t - 2}{19} + 1}{2} - \frac{ln \frac{t - 2}{19} - 1}{2}

Simplify

\frac{1}{19} \frac{ln \frac{t - 2}{19} + 1}{2} - \frac{ln \frac{t - 2}{19} - 1}{2} : \frac{1}{2 19} (ln \frac{t - 2}{19} + 1 - ln \frac{t - 2}{19} - 1)

= \frac{1}{2 19} (ln \frac{t - 2}{19} + 1 - ln \frac{t - 2}{19} - 1)

Add a constant to the solution

= \frac{1}{2 19} (ln \frac{t - 2}{19} + 1 - ln \frac{t - 2}{19} - 1) + C

\frac{1}{1 - x}, 22

\frac{s}{s ^{2} + 2}

f (x) = 4 x - 32 x, at x = 4

x^{2} y^{^{'}} + 2 x y - y^{3} = 0

\int 1 + t^{2} d t

What is the integral of 1/(15+4t-t^2) ?
The integral of 1/(15+4t-t^2) is 1/(2sqrt(19))(ln|(t-2)/(sqrt(19))+1|-ln|(t-2)/(sqrt(19))-1|)+C