What is the integral from 1-2x to 1 of (u^3)/(1+u^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1-2x to 1 of (u^3)/(1+u^2)

\int_{1 - 2 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

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

Solution steps

\int_{1 - 2 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

Apply u-substitution

: \int_{4 x^{2} - 4 x + 2}^{2} \frac{v - 1}{2 v} d v

= \int_{4 x^{2} - 4 x + 2}^{2} \frac{v - 1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{4 x^{2} - 4 x + 2}^{2} \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \frac{1}{2} \cdot \int_{4 x^{2} - 4 x + 2}^{2} 1 - \frac{1}{v} d v

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_{4 x^{2} - 4 x + 2}^{2} 1 d v - \int_{4 x^{2} - 4 x + 2}^{2} \frac{1}{v} d v)

\int_{4 x^{2} - 4 x + 2}^{2} 1 d v = - 4 x^{2} + 4 x

\int_{4 x^{2} - 4 x + 2}^{2} \frac{1}{v} d v = ln (2) - ln 4 x^{2} - 4 x + 2

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

Simplify

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

\int_{2}^{5} 3 x^{2} d x

\int_{- 2}^{- 1} cos (\frac{nπ x}{2}) d x

\int_{0}^{\frac{π}{30}} 20 cos (5 x) d x

\int_{- 3}^{5} (\frac{x}{2} + 5) d x

\int_{- 2}^{2} (6 - x^{2}) d x

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