What is the integral from 0 to 5 of x/(a+bx^2) ?

The integral from 0 to 5 of x/(a+bx^2) is (ln|a+25b|-ln|a|)/(2b)

integral from 0 to 5 of x/(a+bx^2)

\int_{0}^{5} \frac{x}{a + b x ^{2}} d x

\frac{ln ∣ a + 25 b ∣ - ln ∣ a ∣}{2 b}

Solution steps

\int_{0}^{5} \frac{x}{a + b x ^{2}} d x

Apply u-substitution

= \int_{a}^{a + 25 b} \frac{1}{2 b u} d u

Take the constant out:

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

= \frac{1}{2 b} \cdot \int_{a}^{a + 25 b} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2 b} [ln ∣ u ∣]_{a}^{a + 25 b}

Compute the boundaries:

ln ∣ a + 25 b ∣ - ln ∣ a ∣

= \frac{1}{2 b} (ln ∣ a + 25 b ∣ - ln ∣ a ∣)

Simplify

= \frac{ln ∣ a + 25 b ∣ - ln ∣ a ∣}{2 b}

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\int_{2}^{4} 3 x^{2} - 2 d x

What is the integral from 0 to 5 of x/(a+bx^2) ?
The integral from 0 to 5 of x/(a+bx^2) is (ln|a+25b|-ln|a|)/(2b)