What is the integral of (8-3t)/(10t^2+13t-3) ?

The integral of (8-3t)/(10t^2+13t-3) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (8-3t)/(10t^2+13t-3)

\int \frac{8 - 3 t}{10 t ^{2} + 13 t - 3} d t

- \frac{8}{17} (ln \frac{20}{17} t + \frac{30}{17} - ln \frac{20}{17} t - \frac{4}{17}) - 6 (\frac{1}{40} ln 400 t^{2} + 520 t - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} t - ln - \frac{4}{17} + \frac{20}{17} t)) + C

Solution steps

\int \frac{8 - 3 t}{10 t ^{2} + 13 t - 3} d t

Expand

\frac{8 - 3 t}{10 t ^{2} + 13 t - 3} : \frac{8}{10 t ^{2} + 13 t - 3} - \frac{3 t}{10 t ^{2} + 13 t - 3}

Apply the Sum Rule:

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

= \int \frac{8}{10 t ^{2} + 13 t - 3} d t - \int \frac{3 t}{10 t ^{2} + 13 t - 3} d t

\int \frac{8}{10 t ^{2} + 13 t - 3} d t = - \frac{8}{17} (ln \frac{20}{17} t + \frac{30}{17} - ln \frac{20}{17} t - \frac{4}{17})

\int \frac{3 t}{10 t ^{2} + 13 t - 3} d t = 6 (\frac{1}{40} ln 400 t^{2} + 520 t - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} t - ln - \frac{4}{17} + \frac{20}{17} t))

= - \frac{8}{17} (ln \frac{20}{17} t + \frac{30}{17} - ln \frac{20}{17} t - \frac{4}{17}) - 6 (\frac{1}{40} ln 400 t^{2} + 520 t - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} t - ln - \frac{4}{17} + \frac{20}{17} t))

Add a constant to the solution

= - \frac{8}{17} (ln \frac{20}{17} t + \frac{30}{17} - ln \frac{20}{17} t - \frac{4}{17}) - 6 (\frac{1}{40} ln 400 t^{2} + 520 t - 120 + \frac{13}{680} (ln \frac{30}{17} + \frac{20}{17} t - ln - \frac{4}{17} + \frac{20}{17} t)) + C

\int 24 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

\int \frac{y}{y ^{2} + 16} d y

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

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

\int \frac{6}{4 4 x - 1} d x

What is the integral of (8-3t)/(10t^2+13t-3) ?
The integral of (8-3t)/(10t^2+13t-3) is