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Popular >

integral of 4/(x^4+1)

Solution

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

Solution

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 4 \cdot \int \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} + \frac{- x + 2}{2 2 ( x ^{2} - 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

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

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

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

= 4 (\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)))

Simplify

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

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

Add a constant to the solution

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

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Frequently Asked Questions (FAQ)

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