What is the integral from 1 to 3 of (t^6)/(1+t^4) ?

The integral from 1 to 3 of (t^6)/(1+t^4) is

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integral from 1 to 3 of (t^6)/(1+t^4)

\int_{1}^{3} \frac{t ^{6}}{1 + t ^{4}} d t

\frac{26}{3} - \frac{- ln ( 7 - 2 2 ) + 2 arctan ( 3 2 + 1 ) - 2 arctan ( 2 + 1 ) + ln ( 7 + 2 2 ) + 2 arctan ( 3 2 - 1 ) - 2 arctan ( 2 - 1 )}{4 2}

Decimal

8.13220 \dots

Solution steps

\int_{1}^{3} \frac{t ^{6}}{1 + t ^{4}} d t

Long division

\frac{t ^{6}}{1 + t ^{4}} : t^{2} - \frac{t ^{2}}{t ^{4} + 1}

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

\int_{1}^{3} t^{2} d t = \frac{26}{3}

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

= \frac{26}{3} - \frac{- ln ( 7 - 2 2 ) + 2 arctan ( 3 2 + 1 ) - 2 arctan ( 2 + 1 ) + ln ( 7 + 2 2 ) + 2 arctan ( 3 2 - 1 ) - 2 arctan ( 2 - 1 )}{4 2}

What is the integral from 1 to 3 of (t^6)/(1+t^4) ?
The integral from 1 to 3 of (t^6)/(1+t^4) is