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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-(x^2)/(2(x+1))

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

- \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = x + 1

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

Simplify

- \frac{1}{2} (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) : - \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3)

= - \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3)

Add a constant to the solution

= - \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3) + C

\frac{d}{d x} (x + \frac{4}{x - 4})

\int \frac{1}{sin ( 4 x )} d x

2 x^{2} - 8 x - 7

\int + cos^{5} (x) d x

\int \frac{5}{8} csc^{2} (\frac{x}{8}) d x

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