What is the integral from 4 to 3x+4 of t/(1+t^3) ?

The integral from 4 to 3x+4 of t/(1+t^3) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 3x+4 of t/(1+t^3)

\int_{4}^{3 x + 4} \frac{t}{1 + t ^{3}} d t

- \frac{1}{3} (ln ∣ 3 x + 5 ∣ - ln (5)) + \frac{ln 36 x ^{2} + 84 x + 52 - ln ( 52 ) + 2 3 arctan ( \frac{6 x + 7}{3} ) - 2 3 arctan ( \frac{7 3}{3} )}{6}

Solution steps

\int_{4}^{3 x + 4} \frac{t}{1 + t ^{3}} d t

Take the partial fraction of

\frac{t}{1 + t ^{3}} : - \frac{1}{3 ( t + 1 )} + \frac{t + 1}{3 ( t ^{2} - t + 1 )}

= \int_{4}^{3 x + 4} - \frac{1}{3 ( t + 1 )} + \frac{t + 1}{3 ( t ^{2} - t + 1 )} d t

Apply the Sum Rule:

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

= - \int_{4}^{3 x + 4} \frac{1}{3 ( t + 1 )} d t + \int_{4}^{3 x + 4} \frac{t + 1}{3 ( t ^{2} - t + 1 )} d t

\int_{4}^{3 x + 4} \frac{1}{3 ( t + 1 )} d t = \frac{1}{3} (ln ∣ 3 x + 5 ∣ - ln (5))

\int_{4}^{3 x + 4} \frac{t + 1}{3 ( t ^{2} - t + 1 )} d t = \frac{ln 36 x ^{2} + 84 x + 52 - ln ( 52 ) + 2 3 arctan ( \frac{6 x + 7}{3} ) - 2 3 arctan ( \frac{7 3}{3} )}{6}

= - \frac{1}{3} (ln ∣ 3 x + 5 ∣ - ln (5)) + \frac{ln 36 x ^{2} + 84 x + 52 - ln ( 52 ) + 2 3 arctan ( \frac{6 x + 7}{3} ) - 2 3 arctan ( \frac{7 3}{3} )}{6}

\int_{0}^{π} e^{2 x} d x

\int_{- 1}^{0} x^{\frac{1}{3}} d x

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

\int_{1}^{8} (\frac{x}{2} - \frac{9}{x}) d x

\int_{0}^{2} x^{3} x^{4} + 1 d x

What is the integral from 4 to 3x+4 of t/(1+t^3) ?
The integral from 4 to 3x+4 of t/(1+t^3) is