What is the integral from 1 to 4 of (2x+1)/(x^2+x-1) ?

The integral from 1 to 4 of (2x+1)/(x^2+x-1) is ln(19)

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integral from 1 to 4 of (2x+1)/(x^2+x-1)

\int_{1}^{4} \frac{2 x + 1}{x ^{2} + x - 1} d x

ln (19)

Decimal

2.94443 \dots

Solution steps

\int_{1}^{4} \frac{2 x + 1}{x ^{2} + x - 1} d x

Expand

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

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}^{4} \frac{2 x}{x ^{2} + x - 1} d x + \int_{1}^{4} \frac{1}{x ^{2} + x - 1} d x

\int_{1}^{4} \frac{2 x}{x ^{2} + x - 1} d x = ln (19) + \frac{1}{5} ln (\frac{83 - 33 5}{38})

\int_{1}^{4} \frac{1}{x ^{2} + x - 1} d x = - \frac{ln ( \frac{83 - 33 5}{38} )}{5}

= ln (19) + \frac{1}{5} ln (\frac{83 - 33 5}{38}) - \frac{ln ( \frac{83 - 33 5}{38} )}{5}

Simplify

ln (19) + \frac{1}{5} ln (\frac{83 - 33 5}{38}) - \frac{ln ( \frac{83 - 33 5}{38} )}{5} : ln (19)

= ln (19)

\int_{0}^{5} \frac{1}{x - 5} d x

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\int_{0}^{\frac{8}{3}} 3 t - 8 d t

What is the integral from 1 to 4 of (2x+1)/(x^2+x-1) ?
The integral from 1 to 4 of (2x+1)/(x^2+x-1) is ln(19)