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

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

Solution steps

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

\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

\frac{d}{d x} (\frac{1 + x}{x})

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y (0) = 4, y^{^{'}} (0) = 1

f (x) = \frac{9}{x - 6}

x \to 0 lim (\frac{x}{sin ( 6 x )})

f (x) = sin^{2} (e^{s i n^{2} (x)})

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