What is the integral of (-8)/(xsqrt(x+1)) ?

The integral of (-8)/(xsqrt(x+1)) is 16(1/2 ln|sqrt(x+1)+1|-1/2 ln|sqrt(x+1)-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-8)/(xsqrt(x+1))

\int \frac{- 8}{x x + 1} d x

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

Solution steps

\int \frac{- 8}{x x + 1} d x

Take the constant out:

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

= - 8 \cdot \int \frac{1}{x x + 1} d x

Apply u-substitution

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

= - 8 \cdot \int \frac{2}{u ^{2} - 1} d u

Take the constant out:

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

= - 8 \cdot 2 \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= - 8 \cdot 2 \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= - 8 \cdot 2 (- \int \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

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

= - 8 \cdot 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = x + 1

= - 8 \cdot 2 (- (\frac{ln x + 1 + 1}{2} - \frac{ln x + 1 - 1}{2}))

Simplify

- 8 \cdot 2 (- (\frac{ln x + 1 + 1}{2} - \frac{ln x + 1 - 1}{2})) : 16 (\frac{1}{2} ln x + 1 + 1 - \frac{1}{2} ln x + 1 - 1)

= 16 (\frac{1}{2} ln x + 1 + 1 - \frac{1}{2} ln x + 1 - 1)

Add a constant to the solution

= 16 (\frac{1}{2} ln x + 1 + 1 - \frac{1}{2} ln x + 1 - 1) + C

\int (cos (π) x) d x

\int \frac{4}{9 + 2 x ^{2}} d x

\int (3 - 2 x)^{4} d x

\int e^{x} cos (e^{x} + 2) d x

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

What is the integral of (-8)/(xsqrt(x+1)) ?
The integral of (-8)/(xsqrt(x+1)) is 16(1/2 ln|sqrt(x+1)+1|-1/2 ln|sqrt(x+1)-1|)+C