What is the integral from 0 to 3 of (6x)/(1+x^2) ?

The integral from 0 to 3 of (6x)/(1+x^2) is 3ln(10)

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integral from 0 to 3 of (6x)/(1+x^2)

\int_{0}^{3} \frac{6 x}{1 + x ^{2}} d x

3 ln (10)

Decimal

6.90775 \dots

Solution steps

\int_{0}^{3} \frac{6 x}{1 + x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{3} \frac{x}{1 + x ^{2}} d x

Apply u-substitution

: \int_{1}^{10} \frac{1}{2 u} d u

= 6 \cdot \int_{1}^{10} \frac{1}{2 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int_{1}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{10}

Simplify

6 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{10} : 3 [ln ∣ u ∣]_{1}^{10}

= 3 [ln ∣ u ∣]_{1}^{10}

Compute the boundaries:

ln (10)

= 3 ln (10)

\int_{1}^{3} 10 x (x - 1)^{3} d x

\int_{3}^{9} 7 d x

\int_{0}^{1} \frac{2 y}{1 + x ^{2}} d x

\int_{1}^{\infty} x^{\frac{1}{2}} d x

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

What is the integral from 0 to 3 of (6x)/(1+x^2) ?
The integral from 0 to 3 of (6x)/(1+x^2) is 3ln(10)