What is the integral from 1 to 3 of (5x)/(1+x^2) ?

The integral from 1 to 3 of (5x)/(1+x^2) is 5/2 ln(5)

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integral from 1 to 3 of (5x)/(1+x^2)

\int_{1}^{3} \frac{5 x}{1 + x ^{2}} d x

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

Solution steps

\int_{1}^{3} \frac{5 x}{1 + x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{1}^{3} \frac{x}{1 + x ^{2}} d x

Apply u-substitution

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

= 5 \cdot \int_{2}^{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

= 5 \cdot \frac{1}{2} \cdot \int_{2}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{10}

Simplify

5 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{10} : \frac{5}{2} [ln ∣ u ∣]_{2}^{10}

= \frac{5}{2} [ln ∣ u ∣]_{2}^{10}

Compute the boundaries:

ln (10) - ln (2)

= \frac{5}{2} (ln (10) - ln (2))

Simplify

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

\int_{1}^{x^{3}} \frac{1}{1 + ln ( t )} d t

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

\int_{2}^{0} x d x

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

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

What is the integral from 1 to 3 of (5x)/(1+x^2) ?
The integral from 1 to 3 of (5x)/(1+x^2) is 5/2 ln(5)