What is the integral from 1 to sqrt(3 of)arctan(x) ?

The integral from 1 to sqrt(3 of)arctan(x) is (-2sqrt(3)ln(2)+4pi-sqrt(3)pi)/(4sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to sqrt(3 of)arctan(x)

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

\frac{- 2 3 ln ( 2 ) + 4 π - 3 π}{4 3}

Decimal Notation

0.68182 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{- 2 3 ln ( 2 ) + 4 π - 3 π}{4 3}

= \frac{- 2 3 ln ( 2 ) + 4 π - 3 π}{4 3}

\int \frac{x + 1}{x ^{2} + 16} d x

f (x) = x (x - 4)

y^{^{''}} + 2 y^{^{'}} + y = 10 e^{- t}

\frac{d}{d x} (- 2 x e^{- 2 x})

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

What is the integral from 1 to sqrt(3 of)arctan(x) ?
The integral from 1 to sqrt(3 of)arctan(x) is (-2sqrt(3)ln(2)+4pi-sqrt(3)pi)/(4sqrt(3))