What is the integral of (42x^7-7x)/(7x^5) ?

The integral of (42x^7-7x)/(7x^5) is (6x^6+1)/(3x^3)+C

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integral of (42x^7-7x)/(7x^5)

\int \frac{42 x ^{7} - 7 x}{7 x ^{5}} d x

\frac{6 x ^{6} + 1}{3 x ^{3}} + C

Solution steps

\int \frac{42 x ^{7} - 7 x}{7 x ^{5}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int \frac{42 x ^{7} - 7 x}{x ^{5}} d x

Expand

\frac{42 x ^{7} - 7 x}{x ^{5}} : 42 x^{2} - \frac{7}{x ^{4}}

= \frac{1}{7} \cdot \int 42 x^{2} - \frac{7}{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

= \frac{1}{7} (\int 42 x^{2} d x - \int \frac{7}{x ^{4}} d x)

\int 42 x^{2} d x = 14 x^{3}

\int \frac{7}{x ^{4}} d x = - \frac{7}{3 x ^{3}}

= \frac{1}{7} (14 x^{3} - (- \frac{7}{3 x ^{3}}))

Simplify

\frac{1}{7} (14 x^{3} - (- \frac{7}{3 x ^{3}})) : \frac{6 x ^{6} + 1}{3 x ^{3}}

= \frac{6 x ^{6} + 1}{3 x ^{3}}

Add a constant to the solution

= \frac{6 x ^{6} + 1}{3 x ^{3}} + C

\int \frac{1}{2 x ^{6}} d x

f (x) = - x^{2} - 3 x + 4, at x = - 3

\int \frac{x ^{2} - 144}{x} d x

\int cos (y) cos (z) d y

\int_{2}^{3} (\frac{4}{x ^{2}} + 1) d x

What is the integral of (42x^7-7x)/(7x^5) ?
The integral of (42x^7-7x)/(7x^5) is (6x^6+1)/(3x^3)+C