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

The integral from 2 to 3 of x^3arctan(x-1) is -pi+8/3+(-pi+85arctan(2)-24)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 3 of x^3arctan(x-1)

\int_{2}^{3} x^{3} arctan (x - 1) d x

- π + \frac{8}{3} + \frac{- π + 85 arctan ( 2 ) - 24}{4}

Decimal Notation

16.26658 \dots

Solution steps

\int_{2}^{3} x^{3} arctan (x - 1) d x

Apply Integration By Parts

: [\frac{1}{4} x^{4} arctan (x - 1) - \int \frac{x ^{4}}{4 ( x ^{2} - 2 x + 2 )} d x]_{2}^{3}

= [\frac{1}{4} x^{4} arctan (x - 1) - \int \frac{x ^{4}}{4 ( x ^{2} - 2 x + 2 )} d x]_{2}^{3}

\int \frac{x ^{4}}{4 ( x ^{2} - 2 x + 2 )} d x = \frac{x ^{3}}{12} + \frac{1}{4} x^{2} + \frac{1}{2} x - arctan (x - 1)

= [\frac{1}{4} x^{4} arctan (x - 1) - (\frac{x ^{3}}{12} + \frac{1}{4} x^{2} + \frac{1}{2} x - arctan (x - 1))]_{2}^{3}

Simplify

= [\frac{1}{4} x^{4} arctan (x - 1) - \frac{x ^{3}}{12} - \frac{1}{4} x^{2} - \frac{1}{2} x + arctan (x - 1)]_{2}^{3}

Compute the boundaries:

- π + \frac{8}{3} + \frac{- π + 85 arctan ( 2 ) - 24}{4}

= - π + \frac{8}{3} + \frac{- π + 85 arctan ( 2 ) - 24}{4}

\int_{2}^{6} 0.35 e^{- 0.35 x} d x

\int_{1}^{\infty} n e^{- n} d n

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

\int_{0}^{1} (12 ln (x)) d x

\int_{- 1}^{2} (x^{3} - 5 x^{2} + 3 x - 1) d x

What is the integral from 2 to 3 of x^3arctan(x-1) ?
The integral from 2 to 3 of x^3arctan(x-1) is -pi+8/3+(-pi+85arctan(2)-24)/4