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Popular Calculus >

integral of (t^6)/(1+t^4)

Solution

\int \frac{t ^{6}}{1 + t ^{4}} d t

Solution

\frac{t ^{3}}{3} + \frac{1}{4 2} ln 2 t^{2} + 22 t + 2 - \frac{1}{2 2} arctan (2 t + 1) - \frac{1}{4 2} ln 2 t^{2} - 22 t + 2 - \frac{1}{2 2} arctan (2 t - 1) + C

Solution steps

\int \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 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 t^{2} d t - \int \frac{t ^{2}}{t ^{4} + 1} d t

\int t^{2} d t = \frac{t ^{3}}{3}

\int \frac{t ^{2}}{t ^{4} + 1} d t = - \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1))

= \frac{t ^{3}}{3} - (- \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1)))

Simplify

\frac{t ^{3}}{3} - (- \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1))) : \frac{t ^{3}}{3} + \frac{1}{4 2} ln 2 t^{2} + 22 t + 2 - \frac{1}{2 2} arctan (2 t + 1) - \frac{1}{4 2} ln 2 t^{2} - 22 t + 2 - \frac{1}{2 2} arctan (2 t - 1)

= \frac{t ^{3}}{3} + \frac{1}{4 2} ln 2 t^{2} + 22 t + 2 - \frac{1}{2 2} arctan (2 t + 1) - \frac{1}{4 2} ln 2 t^{2} - 22 t + 2 - \frac{1}{2 2} arctan (2 t - 1)

Add a constant to the solution

= \frac{t ^{3}}{3} + \frac{1}{4 2} ln 2 t^{2} + 22 t + 2 - \frac{1}{2 2} arctan (2 t + 1) - \frac{1}{4 2} ln 2 t^{2} - 22 t + 2 - \frac{1}{2 2} arctan (2 t - 1) + C

Graph

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Frequently Asked Questions (FAQ)

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