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

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

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

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

ln b^{2} + 4 - ln (5)

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{5}^{b^{2} + 4} \frac{1}{2 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

= [ln ∣ u ∣]_{5}^{b^{2} + 4}

Compute the boundaries:

ln b^{2} + 4 - ln (5)

= ln b^{2} + 4 - ln (5)

x^{2} y + y^{^{'}} = - 3 x^{2}

f (x) = 1 + 4 x, at x = 2

x \to c lim (x^{3} + 4 x^{2} - 3)

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

\frac{- x ^{2} - 7}{( x ^{2} - 7 ) ^{2}}

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