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

The integral from 1 to 2 of x/(1+4x^2) is 1/8 (ln(17)-ln(5))

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

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

\frac{1}{8} (ln (17) - ln (5))

Decimal

0.15297 \dots

Solution steps

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

Apply u-substitution

: \int_{5}^{17} \frac{1}{8 u} d u

= \int_{5}^{17} \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{5}^{17} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} [ln ∣ u ∣]_{5}^{17}

Compute the boundaries:

ln (17) - ln (5)

= \frac{1}{8} (ln (17) - ln (5))

(- \frac{3}{7}, 0), (0, 6)

\int_{0}^{5} e^{x} d x

x \to 4 lim (3 x^{5} - 4)

x, x - 2, 0, 4

\frac{s}{( s + 3 ) ^{2} + 1}

What is the integral from 1 to 2 of x/(1+4x^2) ?
The integral from 1 to 2 of x/(1+4x^2) is 1/8 (ln(17)-ln(5))