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

The integral of (sqrt(x)-x+1)/(x^5) is -2/(7x^{7/2)}+1/(3x^3)-1/(4x^4)+C

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

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

- \frac{2}{7 x ^{\frac{7}{2}}} + \frac{1}{3 x ^{3}} - \frac{1}{4 x ^{4}} + C

Solution steps

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

Expand

\frac{x - x + 1}{x ^{5}} : \frac{1}{x ^{\frac{9}{2}}} - \frac{1}{x ^{4}} + \frac{1}{x ^{5}}

= \int \frac{1}{x ^{\frac{9}{2}}} - \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 ^{\frac{9}{2}}} d x - \int \frac{1}{x ^{4}} d x + \int \frac{1}{x ^{5}} d x

\int \frac{1}{x ^{\frac{9}{2}}} d x = - \frac{2}{7 x ^{\frac{7}{2}}}

\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{2}{7 x ^{\frac{7}{2}}} - (- \frac{1}{3 x ^{3}}) - \frac{1}{4 x ^{4}}

Simplify

= - \frac{2}{7 x ^{\frac{7}{2}}} + \frac{1}{3 x ^{3}} - \frac{1}{4 x ^{4}}

Add a constant to the solution

= - \frac{2}{7 x ^{\frac{7}{2}}} + \frac{1}{3 x ^{3}} - \frac{1}{4 x ^{4}} + C

\int_{- \infty}^{\infty} x^{8} e^{- x^{9}} d x

\int - 4 x d x

\int - cos (x^{2}) d x

\frac{d}{d x} (x 3 + 2 x^{2})

(\frac{1}{ln ( x )})^{^{'}}

What is the integral of (sqrt(x)-x+1)/(x^5) ?
The integral of (sqrt(x)-x+1)/(x^5) is -2/(7x^{7/2)}+1/(3x^3)-1/(4x^4)+C