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

The integral of 1/(x^4)+4 is -1/(3x^3)+4x+C

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integral of 1/(x^4)+4

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

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

Solution steps

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

Add a constant to the solution

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

\frac{d y}{d x}, x^{3} + y^{3} - 9 x y = 0

f (t) = 1 + e^{4 t}

y = 2 x, (18, 6)

\frac{\partial}{\partial x} (\frac{y e ^{- \frac{y}{x}}}{x ^{2}})

\int - 6 y d y