What is the integral of (5x+4)/(x^3sqrt(x^2+1)) ?

The integral of (5x+4)/(x^3sqrt(x^2+1)) is

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integral of (5x+4)/(x^3sqrt(x^2+1))

\int \frac{5 x + 4}{x ^{3} x ^{2} + 1} d x

- \frac{5 1 + x ^{2}}{x} + \frac{1}{2} tan^{2} (\frac{arctan ( x )}{2}) - 2 ln tan (\frac{1}{2} arctan (x)) - \frac{1}{2} cot^{2} (\frac{arctan ( x )}{2}) + C

Solution steps

\int \frac{5 x + 4}{x ^{3} x ^{2} + 1} d x

Expand

\frac{5 x + 4}{x ^{3} x ^{2} + 1} : \frac{5}{x ^{2} x ^{2} + 1} + \frac{4}{x ^{3} x ^{2} + 1}

= \int \frac{5}{x ^{2} x ^{2} + 1} + \frac{4}{x ^{3} x ^{2} + 1} 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{5}{x ^{2} x ^{2} + 1} d x + \int \frac{4}{x ^{3} x ^{2} + 1} d x

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

\int \frac{4}{x ^{3} x ^{2} + 1} d x = \frac{1}{2} tan^{2} (\frac{arctan ( x )}{2}) - 2 ln tan (\frac{1}{2} arctan (x)) - \frac{1}{2} cot^{2} (\frac{arctan ( x )}{2})

= - \frac{5 1 + x ^{2}}{x} + \frac{1}{2} tan^{2} (\frac{arctan ( x )}{2}) - 2 ln tan (\frac{1}{2} arctan (x)) - \frac{1}{2} cot^{2} (\frac{arctan ( x )}{2})

Add a constant to the solution

= - \frac{5 1 + x ^{2}}{x} + \frac{1}{2} tan^{2} (\frac{arctan ( x )}{2}) - 2 ln tan (\frac{1}{2} arctan (x)) - \frac{1}{2} cot^{2} (\frac{arctan ( x )}{2}) + C

What is the integral of (5x+4)/(x^3sqrt(x^2+1)) ?
The integral of (5x+4)/(x^3sqrt(x^2+1)) is