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Exercise

$\int\frac{1}{\sqrt{16-x^4}}-\frac{1}{4}dx$

Step-by-step Solution

1

Expand the integral $\int\left(\frac{1}{\sqrt{16-x^4}}-\frac{1}{4}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{1}{\sqrt{16-x^4}}dx+\int-\frac{1}{4}dx$
2

The integral $\int\frac{1}{\sqrt{16-x^4}}dx$ results in: $-\frac{1}{2}F\left(\frac{\arcsin\left(\frac{\sqrt{16-x^4}}{4}\right)}{2}\Big\vert 2\right)$

$-\frac{1}{2}F\left(\frac{\arcsin\left(\frac{\sqrt{16-x^4}}{4}\right)}{2}\Big\vert 2\right)$
3

The integral $\int-\frac{1}{4}dx$ results in: $-\frac{1}{4}x$

$-\frac{1}{4}x$
4

Gather the results of all integrals

$-\frac{1}{2}F\left(\frac{\arcsin\left(\frac{\sqrt{16-x^4}}{4}\right)}{2}\Big\vert 2\right)-\frac{1}{4}x$
5

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$-\frac{1}{2}F\left(\frac{\arcsin\left(\frac{\sqrt{16-x^4}}{4}\right)}{2}\Big\vert 2\right)-\frac{1}{4}x+C_0$

Final answer to the exercise

$-\frac{1}{2}F\left(\frac{\arcsin\left(\frac{\sqrt{16-x^4}}{4}\right)}{2}\Big\vert 2\right)-\frac{1}{4}x+C_0$

Try other ways to solve this exercise

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