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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 (ln (5) - ln (2))

Decimal

1.83258 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \frac{1}{2} \cdot \int_{2}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{5}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{5} : 2 [ln ∣ u ∣]_{2}^{5}

= 2 [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

= 2 (ln (5) - ln (2))

\int_{0}^{1} x e^{x + 2} d x

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

\int_{1}^{2} x e^{5 x^{2}} d x

\int_{0}^{1} k (x^{3} + x^{2}) d x

\int_{- t}^{t} 1 - x^{2} - y^{2} d y

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