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

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

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

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

\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln (5) + \frac{1}{6} (- ln \frac{( 2 x - 1 ) ^{2}}{3} + 1 + ln (\frac{52}{3}) + 23 arctan (\frac{2 x - 1}{3}) - 23 arctan (\frac{7}{3}))

Solution steps

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

Take the partial fraction of

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

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

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

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

\int_{4}^{x} \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x = \frac{1}{6} (- ln \frac{( 2 x - 1 ) ^{2}}{3} + 1 + ln (\frac{52}{3}) + 23 arctan (\frac{2 x - 1}{3}) - 23 arctan (\frac{7 3}{3}))

= \frac{1}{3} (ln ∣ x + 1 ∣ - ln (5)) + \frac{1}{6} (- ln \frac{( 2 x - 1 ) ^{2}}{3} + 1 + ln (\frac{52}{3}) + 23 arctan (\frac{2 x - 1}{3}) - 23 arctan (\frac{7 3}{3}))

Simplify

\frac{1}{3} (ln ∣ x + 1 ∣ - ln (5)) + \frac{1}{6} (- ln \frac{( 2 x - 1 ) ^{2}}{3} + 1 + ln (\frac{52}{3}) + 23 arctan (\frac{2 x - 1}{3}) - 23 arctan (\frac{7 3}{3})) : \frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln (5) + \frac{1}{6} (- ln \frac{( 2 x - 1 ) ^{2}}{3} + 1 + ln (\frac{52}{3}) + 23 arctan (\frac{2 x - 1}{3}) - 23 arctan (\frac{7}{3}))

= \frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln (5) + \frac{1}{6} (- ln \frac{( 2 x - 1 ) ^{2}}{3} + 1 + ln (\frac{52}{3}) + 23 arctan (\frac{2 x - 1}{3}) - 23 arctan (\frac{7}{3}))

\int_{0}^{\infty} e^{c t} d t

\int_{0}^{5} (\frac{x ^{2}}{3} + 1) d x

\int_{- 1}^{1} \frac{x + 2}{56} d x

\int_{1}^{\infty} c e^{- x} d x

\int_{5}^{6} x (x^{2} - 25)^{2} d x

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