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

The integral from 1 to 2 of 1/(x^3+4x) is (-ln(2)+ln(5))/8

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integral from 1 to 2 of 1/(x^3+4x)

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

\frac{- ln ( 2 ) + ln ( 5 )}{8}

Decimal Notation

0.11453 \dots

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{3} + 4 x} : \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

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

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

\int_{1}^{2} \frac{1}{4 x} d x = \frac{1}{4} ln (2)

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= \frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5))

Simplify

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)) : \frac{- ln ( 2 ) + ln ( 5 )}{8}

= \frac{- ln ( 2 ) + ln ( 5 )}{8}

(5 e^{9 t} - 4) e^{- 6 t}

(t)^{2}

\int \frac{4}{81 - ( x + 9 ) ^{2}} d x

y^{^{'}} + 7 y = 3

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

What is the integral from 1 to 2 of 1/(x^3+4x) ?
The integral from 1 to 2 of 1/(x^3+4x) is (-ln(2)+ln(5))/8