What is the integral of x^{-6}(x+1) ?

The integral of x^{-6}(x+1) is -1/(4x^4)-1/(5x^5)+C

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integral of x^{-6}(x+1)

\int x^{- 6} (x + 1) d x

- \frac{1}{4 x ^{4}} - \frac{1}{5 x ^{5}} + C

Solution steps

\int x^{- 6} (x + 1) d x

Simplify

x^{- 6} (x + 1) : \frac{x + 1}{x ^{6}}

= \int \frac{x + 1}{x ^{6}} d x

Expand

\frac{x + 1}{x ^{6}} : \frac{1}{x ^{5}} + \frac{1}{x ^{6}}

= \int \frac{1}{x ^{5}} + \frac{1}{x ^{6}} 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 ^{5}} d x + \int \frac{1}{x ^{6}} d x

\int \frac{1}{x ^{5}} d x = - \frac{1}{4 x ^{4}}

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

= - \frac{1}{4 x ^{4}} - \frac{1}{5 x ^{5}}

Add a constant to the solution

= - \frac{1}{4 x ^{4}} - \frac{1}{5 x ^{5}} + C

y^{^{''}} - 9 y^{^{'}} + 18 y = 2 sin (5 t)

e^{c o s (s i n (x^{2}))}

\int_{- 1}^{1} - 1 d x

\int (x - \frac{1}{x})^{2} d x

\frac{\partial}{\partial x} (e^{- 1})

What is the integral of x^{-6}(x+1) ?
The integral of x^{-6}(x+1) is -1/(4x^4)-1/(5x^5)+C