What is the integral of x^{-4}(x+1) ?

The integral of x^{-4}(x+1) is -1/(2x^2)-1/(3x^3)+C

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integral of x^{-4}(x+1)

\int x^{- 4} (x + 1) d x

- \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} + C

Solution steps

\int x^{- 4} (x + 1) d x

Simplify

x^{- 4} (x + 1) : \frac{x + 1}{x ^{4}}

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

Expand

\frac{x + 1}{x ^{4}} : \frac{1}{x ^{3}} + \frac{1}{x ^{4}}

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

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

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

= - \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}}

Add a constant to the solution

= - \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} + C

\frac{d}{d x} ((x^{2} - 2)^{2})

\int_{1}^{3} \frac{6}{x} d x

f (x) = 4 x sin (x) cos (x)

(- 3, - 10), (4, 5)

y = 5 x^{4} f (x)

What is the integral of x^{-4}(x+1) ?
The integral of x^{-4}(x+1) is -1/(2x^2)-1/(3x^3)+C