What is the integral of 1/(x^4)-1/(\sqrt[4]{x)}-4 ?

The integral of 1/(x^4)-1/(\sqrt[4]{x)}-4 is -1/(3x^3)-4/3 x^{3/4}-4x+C

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integral of 1/(x^4)-1/(\sqrt[4]{x)}-4

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

- \frac{1}{3 x ^{3}} - \frac{4}{3} x^{\frac{3}{4}} - 4 x + C

Solution steps

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

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

\int \frac{1}{4 x} d x = \frac{4}{3} x^{\frac{3}{4}}

\int 4 d x = 4 x

= - \frac{1}{3 x ^{3}} - \frac{4}{3} x^{\frac{3}{4}} - 4 x

Add a constant to the solution

= - \frac{1}{3 x ^{3}} - \frac{4}{3} x^{\frac{3}{4}} - 4 x + C

f (x) = sec^{2} (π x)

\int - \frac{1}{4} x d x

e^{- 2 t} cos (3 t)

\frac{\partial}{\partial y} (y + 1)

\frac{6}{( s + 1 ) ( 2 s ^{2} + 5 s + 3 )}

What is the integral of 1/(x^4)-1/(\sqrt[4]{x)}-4 ?
The integral of 1/(x^4)-1/(\sqrt[4]{x)}-4 is -1/(3x^3)-4/3 x^{3/4}-4x+C