What is the integral of ((x^2-x+42))/(x^3+7x) ?

The integral of ((x^2-x+42))/(x^3+7x) is 6ln|x|-5/2 ln|x^2+7|-1/(sqrt(7))arctan(x/(sqrt(7)))+C

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integral of ((x^2-x+42))/(x^3+7x)

\int \frac{( x ^{2} - x + 42 )}{x ^{3} + 7 x} d x

6 ln ∣ x ∣ - \frac{5}{2} ln x^{2} + 7 - \frac{1}{7} arctan (\frac{x}{7}) + C

Solution steps

\int \frac{x ^{2} - x + 42}{x ^{3} + 7 x} d x

Take the partial fraction of

\frac{x ^{2} - x + 42}{x ^{3} + 7 x} : \frac{6}{x} + \frac{- 5 x - 1}{x ^{2} + 7}

= \int \frac{6}{x} + \frac{- 5 x - 1}{x ^{2} + 7} 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{6}{x} d x + \int \frac{- 5 x - 1}{x ^{2} + 7} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{- 5 x - 1}{x ^{2} + 7} d x = - \frac{5}{2} ln x^{2} + 7 - \frac{1}{7} arctan (\frac{x}{7})

= 6 ln ∣ x ∣ - \frac{5}{2} ln x^{2} + 7 - \frac{1}{7} arctan (\frac{x}{7})

Add a constant to the solution

= 6 ln ∣ x ∣ - \frac{5}{2} ln x^{2} + 7 - \frac{1}{7} arctan (\frac{x}{7}) + C

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

f (x) = (\frac{x + 3}{x - 3})^{5}

\int (x + \frac{1}{x ^{3}}) d x

n = 1 \sum \infty \frac{1}{2} (0.3)^{n}

f (x) = \frac{3}{x}, at x = - 5

What is the integral of ((x^2-x+42))/(x^3+7x) ?
The integral of ((x^2-x+42))/(x^3+7x) is 6ln|x|-5/2 ln|x^2+7|-1/(sqrt(7))arctan(x/(sqrt(7)))+C