What is the integral of (x^7+x^2+1)/(x^2) ?

The integral of (x^7+x^2+1)/(x^2) is (x^6)/6+x-1/x+C

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integral of (x^7+x^2+1)/(x^2)

\int \frac{x ^{7} + x ^{2} + 1}{x ^{2}} d x

\frac{x ^{6}}{6} + x - \frac{1}{x} + C

Solution steps

\int \frac{x ^{7} + x ^{2} + 1}{x ^{2}} d x

Expand

\frac{x ^{7} + x ^{2} + 1}{x ^{2}} : x^{5} + 1 + \frac{1}{x ^{2}}

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

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

\int 1 d x = x

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

= \frac{x ^{6}}{6} + x - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{6}}{6} + x - \frac{1}{x} + C

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What is the integral of (x^7+x^2+1)/(x^2) ?
The integral of (x^7+x^2+1)/(x^2) is (x^6)/6+x-1/x+C