What is the integral of 1/(2-3u-u^2) ?

The integral of 1/(2-3u-u^2) is 1/(sqrt(17))(ln|(2u+3)/(sqrt(17))+1|-ln|(2u+3)/(sqrt(17))-1|)+C

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integral of 1/(2-3u-u^2)

\int \frac{1}{2 - 3 u - u ^{2}} d u

\frac{1}{17} (ln \frac{2 u + 3}{17} + 1 - ln \frac{2 u + 3}{17} - 1) + C

Solution steps

\int \frac{1}{2 - 3 u - u ^{2}} d u

Complete the square

2 - 3 u - u^{2} : - (u + \frac{3}{2})^{2} + \frac{17}{4}

= \int \frac{1}{- ( u + \frac{3}{2} ) ^{2} + \frac{17}{4}} d u

Apply u-substitution

: \int \frac{4}{- 4 v ^{2} + 17} d v

= \int \frac{4}{- 4 v ^{2} + 17} d v

Take the constant out:

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

= 4 \cdot \int \frac{1}{- 4 v ^{2} + 17} d v

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 17 ( - w ^{2} + 1 )} d w

Take the constant out:

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

= 4 \cdot \frac{1}{2 17} \cdot \int \frac{1}{- w ^{2} + 1} d w

Use the common integral:

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

= 4 \cdot \frac{1}{2 17} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Substitute back

= 4 \cdot \frac{1}{2 17} \frac{ln \frac{2}{17} ( u + \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{17} ( u + \frac{3}{2} ) - 1}{2}

Simplify

4 \cdot \frac{1}{2 17} \frac{ln \frac{2}{17} ( u + \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{17} ( u + \frac{3}{2} ) - 1}{2} : \frac{1}{17} (ln \frac{2 u + 3}{17} + 1 - ln \frac{2 u + 3}{17} - 1)

= \frac{1}{17} (ln \frac{2 u + 3}{17} + 1 - ln \frac{2 u + 3}{17} - 1)

Add a constant to the solution

= \frac{1}{17} (ln \frac{2 u + 3}{17} + 1 - ln \frac{2 u + 3}{17} - 1) + C

\frac{d}{d x} (lo g_{2} (x + 1))

7 x^{2}

(x^{4})^{^{'}}

\frac{\partial}{\partial x} (\frac{1}{2} y arctan (\frac{x}{y}))

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 )} d x

What is the integral of 1/(2-3u-u^2) ?
The integral of 1/(2-3u-u^2) is 1/(sqrt(17))(ln|(2u+3)/(sqrt(17))+1|-ln|(2u+3)/(sqrt(17))-1|)+C