What is the integral of (x+1)/(x^5) ?

The integral of (x+1)/(x^5) is -1/(3x^3)-1/(4x^4)+C

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integral of (x+1)/(x^5)

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

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

Solution steps

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

Expand

\frac{x + 1}{x ^{5}} : \frac{1}{x ^{4}} + \frac{1}{x ^{5}}

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

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

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

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

Add a constant to the solution

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

y = 9 arctan (x + 1 + x^{2})

\frac{d y}{d x}, y = x^{2} + 5

- 1.8 t^{2} + 14.4 t

y = t sin (t) + cos (t)

\int 9 ln (x) d x

What is the integral of (x+1)/(x^5) ?
The integral of (x+1)/(x^5) is -1/(3x^3)-1/(4x^4)+C