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

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

Popular >

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

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

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

Decimal Notation

0.45814 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{5} \frac{1}{2 u} d u

= \int_{2}^{5} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

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

\int_{0}^{1} 2 t d t

\int_{0}^{\frac{π}{3}} sin^{6} (x) d x

\int_{- 1}^{2} t sin (t^{2}) + 2 t^{3} d t

\int_{1}^{4} 9 x^{2} d x

\int_{- 2}^{5} - \frac{5}{x ^{2}} d x

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