What is the integral of 4/(5+2x-x^2) ?

The integral of 4/(5+2x-x^2) is (sqrt(2))/(sqrt(3))(ln|(x-1)/(sqrt(6))+1|-ln|(x-1)/(sqrt(6))-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4/(5+2x-x^2)

\int \frac{4}{5 + 2 x - x ^{2}} d x

\frac{2}{3} (ln \frac{x - 1}{6} + 1 - ln \frac{x - 1}{6} - 1) + C

Solution steps

\int \frac{4}{5 + 2 x - x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{5 + 2 x - x ^{2}} d x

Complete the square

5 + 2 x - x^{2} : - (x - 1)^{2} + 6

= 4 \cdot \int \frac{1}{- ( x - 1 ) ^{2} + 6} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 6} d u

= 4 \cdot \int \frac{1}{- u ^{2} + 6} d u

Apply Integral Substitution

= 4 \cdot \int \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

= 4 \cdot \frac{1}{6} \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}

= 4 \cdot \frac{1}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 4 \cdot \frac{1}{6} \frac{ln \frac{x - 1}{6} + 1}{2} - \frac{ln \frac{x - 1}{6} - 1}{2}

Simplify

4 \cdot \frac{1}{6} \frac{ln \frac{x - 1}{6} + 1}{2} - \frac{ln \frac{x - 1}{6} - 1}{2} : \frac{2}{3} (ln \frac{x - 1}{6} + 1 - ln \frac{x - 1}{6} - 1)

= \frac{2}{3} (ln \frac{x - 1}{6} + 1 - ln \frac{x - 1}{6} - 1)

Add a constant to the solution

= \frac{2}{3} (ln \frac{x - 1}{6} + 1 - ln \frac{x - 1}{6} - 1) + C

3 (s^{3} - s^{2})

\frac{d}{d x} (e^{8 x s i n (2 x)})

\frac{d}{d x} (\frac{2 x}{9 x ^{2} + 1})

\int \frac{1}{4 x ^{2} + 2 x + 1} d x

\frac{1}{u}

What is the integral of 4/(5+2x-x^2) ?
The integral of 4/(5+2x-x^2) is (sqrt(2))/(sqrt(3))(ln|(x-1)/(sqrt(6))+1|-ln|(x-1)/(sqrt(6))-1|)+C