What is the integral from 1 to x^2 of (t-1)/(t^3+1) ?

The integral from 1 to x^2 of (t-1)/(t^3+1) is -2/3 (ln|x^2+1|-ln(2))+1/3 ln|x^4-x^2+1|

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integral from 1 to x^2 of (t-1)/(t^3+1)

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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What is the integral from 1 to x^2 of (t-1)/(t^3+1) ?
The integral from 1 to x^2 of (t-1)/(t^3+1) is -2/3 (ln|x^2+1|-ln(2))+1/3 ln|x^4-x^2+1|