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Popular >

integral of (x^4)/(1+x^4)

Solution

\int \frac{x ^{4}}{1 + x ^{4}} d x

Solution

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

Solution steps

\int \frac{x ^{4}}{1 + x ^{4}} d x

\frac{x ^{4}}{1 + x ^{4}} = - \frac{1}{1 + x ^{4}} + 1

= \int - \frac{1}{1 + x ^{4}} + 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 \frac{1}{1 + x ^{4}} d x + \int 1 d x

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

\int 1 d x = x

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

Simplify

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

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

Add a constant to the solution

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

Graph

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Frequently Asked Questions (FAQ)

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