What is the integral from 1 to 5 of (x-1)/(x^3+x) ?

The integral from 1 to 5 of (x-1)/(x^3+x) is -ln(5)+1/2 ln(13)+arctan(5)-pi/4

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integral from 1 to 5 of (x-1)/(x^3+x)

\int_{1}^{5} \frac{x - 1}{x ^{3} + x} d x

- ln (5) + \frac{1}{2} ln (13) + arctan (5) - \frac{π}{4}

Decimal Notation

0.26103 \dots

Solution steps

\int_{1}^{5} \frac{x - 1}{x ^{3} + x} d x

Take the partial fraction of

\frac{x - 1}{x ^{3} + x} : - \frac{1}{x} + \frac{x + 1}{x ^{2} + 1}

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

\int_{1}^{5} \frac{1}{x} d x = ln (5)

\int_{1}^{5} \frac{x + 1}{x ^{2} + 1} d x = \frac{1}{2} ln (13) + arctan (5) - \frac{π}{4}

= - ln (5) + \frac{1}{2} ln (13) + arctan (5) - \frac{π}{4}

9 y^{^{'}} + y = 5 t^{2}

n = 1 \sum \infty (- \frac{3}{5})^{n}

\frac{d}{d u} (tan (\frac{u}{v}))

\frac{d x}{d y} = y

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

What is the integral from 1 to 5 of (x-1)/(x^3+x) ?
The integral from 1 to 5 of (x-1)/(x^3+x) is -ln(5)+1/2 ln(13)+arctan(5)-pi/4