What is the integral from 0 to 1 of arctan(5x+3) ?

The integral from 0 to 1 of arctan(5x+3) is (ln(10)-ln(65)-6arctan(3)+16arctan(8))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of arctan(5x+3)

\int_{0}^{1} arctan (5 x + 3) d x

\frac{ln ( 10 ) - ln ( 65 ) - 6 arctan ( 3 ) + 16 arctan ( 8 )}{10}

Decimal

1.37769 \dots

Solution steps

\int_{0}^{1} arctan (5 x + 3) d x

Multiply by the conjugate of

5 x + 3 : \frac{25 x ^{2} - 9}{5 x - 3}

= \int_{0}^{1} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x

Find undefined (singularity) points:

x = \frac{3}{5}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{3}{5}} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x + \int_{\frac{3}{5}}^{1} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x

\int_{0}^{\frac{3}{5}} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x = \frac{12 arctan ( 6 ) - ln ( 37 ) - 6 arctan ( 3 ) + ln ( 10 )}{10}

\int_{\frac{3}{5}}^{1} arctan (\frac{25 x ^{2} - 9}{5 x - 3}) d x = \frac{8}{5} arctan (8) - \frac{1}{10} ln (65) - \frac{6}{5} arctan (6) + \frac{1}{10} ln (37)

= \frac{12 arctan ( 6 ) - ln ( 37 ) - 6 arctan ( 3 ) + ln ( 10 )}{10} + \frac{8}{5} arctan (8) - \frac{1}{10} ln (65) - \frac{6}{5} arctan (6) + \frac{1}{10} ln (37)

Simplify

\frac{12 arctan ( 6 ) - ln ( 37 ) - 6 arctan ( 3 ) + ln ( 10 )}{10} + \frac{8}{5} arctan (8) - \frac{1}{10} ln (65) - \frac{6}{5} arctan (6) + \frac{1}{10} ln (37) : \frac{ln ( 10 ) - ln ( 65 ) - 6 arctan ( 3 ) + 16 arctan ( 8 )}{10}

= \frac{ln ( 10 ) - ln ( 65 ) - 6 arctan ( 3 ) + 16 arctan ( 8 )}{10}

t an g e n t f (x) = \frac{2 x}{x + 2}, (2, 1)

in v erse l a pl a ce \frac{1}{( ( s + 1 ) ^{2} + 4 )}

\int sin^{5} (t co) s^{4} t d t

\frac{d}{d x} (, x e^{y})

t an g e n t ln (x^{2} - 24) = x - y + 6

What is the integral from 0 to 1 of arctan(5x+3) ?
The integral from 0 to 1 of arctan(5x+3) is (ln(10)-ln(65)-6arctan(3)+16arctan(8))/(10)