What is the integral of 1/(x^2+b^4) ?

The integral of 1/(x^2+b^4) is 1/(b^2)arctan(x/(b^2))+C

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integral of 1/(x^2+b^4)

\int \frac{1}{x ^{2} + b ^{4}} d x

\frac{1}{b ^{2}} arctan (\frac{x}{b ^{2}}) + C

Solution steps

\int \frac{1}{x ^{2} + b ^{4}} d x

Apply Integral Substitution

= \int \frac{1}{b ^{2} ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{b ^{2}} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{b ^{2}} arctan (u)

Substitute back

u = \frac{x}{b ^{2}}

= \frac{1}{b ^{2}} arctan (\frac{x}{b ^{2}})

Add a constant to the solution

= \frac{1}{b ^{2}} arctan (\frac{x}{b ^{2}}) + C

\int \frac{6}{49 - x ^{2}} d x

\int \frac{arcsin ( \frac{1}{x} )}{x ^{5}} d x

\int \frac{- 5}{2 x} d x

\int \frac{x ^{2}}{( 49 + x ^{2} ) ^{2}} d x

What is the integral of 1/(x^2+b^4) ?
The integral of 1/(x^2+b^4) is 1/(b^2)arctan(x/(b^2))+C