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Popular >

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

Solution

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

Solution

\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)) + C

Solution steps

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

Take the partial fraction of

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

= \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

= \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))

= \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))

Add a constant to the solution

= \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)) + C

Popular Examples

integral of (x^4)/((7+x^5)^2)

\int \frac{x ^{4}}{( 7 + x ^{5} ) ^{2}} d x

sum from n=2 to infinity of (3^n)/(12^n)

n = 2 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

4y^{''}-4y^'+y=8e^{x/2}

4 y^{^{''}} - 4 y^{^{'}} + y = 8 e^{\frac{x}{2}}

derivative of y=8e^{17t}derivative of

y = 8 e^{17 t}

derivative of 4csc(5x)derivative of

4 csc (5 x)

Frequently Asked Questions (FAQ)

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