What is the integral from 0 to 1 of (x^2)/(x^3-5) ?

The integral from 0 to 1 of (x^2)/(x^3-5) is 1/3 (2ln(2)-ln(5))

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integral from 0 to 1 of (x^2)/(x^3-5)

\int_{0}^{1} \frac{x ^{2}}{x ^{3} - 5} d x

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

Decimal Notation

- 0.07438 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{x ^{3} - 5} d x

Apply u-substitution

: \int_{- 5}^{- 4} \frac{1}{3 u} d u

= \int_{- 5}^{- 4} \frac{1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{- 5}^{- 4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{- 5}^{- 4}

Compute the boundaries:

2 ln (2) - ln (5)

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

\int \frac{x ^{3}}{x ^{2} + 121} d x

f (x) = \frac{1}{x ^{2} - 5}

\int sin (22 x) sin (23 x) d x

\int \frac{- 1}{t} d t

x \to 1 + lim (x^{\frac{x}{1 - x}})

What is the integral from 0 to 1 of (x^2)/(x^3-5) ?
The integral from 0 to 1 of (x^2)/(x^3-5) is 1/3 (2ln(2)-ln(5))