What is the integral from 3 to 3x+5 of t/(1+t^3) ?

The integral from 3 to 3x+5 of t/(1+t^3) is

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integral from 3 to 3x+5 of t/(1+t^3)

\int_{3}^{3 x + 5} \frac{t}{1 + t ^{3}} d t

- \frac{1}{3} ln ∣ 3 x + 6 ∣ + \frac{2}{3} ln (2) + \frac{ln 36 x ^{2} + 108 x + 84 - ln ( 28 ) + 2 3 arctan ( 3 ( 2 x + 3 ) ) - 2 3 arctan ( \frac{5}{3} )}{6}

Solution steps

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

\int_{3}^{3 x + 5} \frac{1}{3 ( t + 1 )} d t = \frac{1}{3} (ln ∣ 3 x + 6 ∣ - 2 ln (2))

\int_{3}^{3 x + 5} \frac{t + 1}{3 ( t ^{2} - t + 1 )} d t = \frac{ln 36 x ^{2} + 108 x + 84 - ln ( 28 ) + 2 3 arctan ( 3 ( 2 x + 3 ) ) - 2 3 arctan ( \frac{5 3}{3} )}{6}

= - \frac{1}{3} (ln ∣ 3 x + 6 ∣ - 2 ln (2)) + \frac{ln 36 x ^{2} + 108 x + 84 - ln ( 28 ) + 2 3 arctan ( 3 ( 2 x + 3 ) ) - 2 3 arctan ( \frac{5 3}{3} )}{6}

Simplify

- \frac{1}{3} (ln ∣ 3 x + 6 ∣ - 2 ln (2)) + \frac{ln 36 x ^{2} + 108 x + 84 - ln ( 28 ) + 2 3 arctan ( 3 ( 2 x + 3 ) ) - 2 3 arctan ( \frac{5 3}{3} )}{6} : - \frac{1}{3} ln ∣ 3 x + 6 ∣ + \frac{2}{3} ln (2) + \frac{ln 36 x ^{2} + 108 x + 84 - ln ( 28 ) + 2 3 arctan ( 3 ( 2 x + 3 ) ) - 2 3 arctan ( \frac{5}{3} )}{6}

= - \frac{1}{3} ln ∣ 3 x + 6 ∣ + \frac{2}{3} ln (2) + \frac{ln 36 x ^{2} + 108 x + 84 - ln ( 28 ) + 2 3 arctan ( 3 ( 2 x + 3 ) ) - 2 3 arctan ( \frac{5}{3} )}{6}

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What is the integral from 3 to 3x+5 of t/(1+t^3) ?
The integral from 3 to 3x+5 of t/(1+t^3) is