integral from -1 to 3 of x/(x^2+4) — step-by-step solution

1) Let's compute the integral:

\int_{-1}^{3} \frac{x}{x^{2} + 4} d x

For the math expression:

\frac{x}{x^{2} + 4}

, since (ln(a))' = a'/a, we try differentiating

\ln (x^{2} + 4)

2) Let's differentiate:

\frac{d}{d x} (\ln (x^{2} + 4))

We differentiate using:

\frac{d}{d x} (\ln (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f}

(where f =

x^{2} + 4

)

ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d x} (\ln (x^{2} + 4))

\frac{d}{d x} (x^{2} + 4) \cdot \frac{1}{x^{2} + 4}

\frac{d}{d x} (x^{2} + 4) \cdot \frac{1}{x^{2} + 4}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{d}{d x} (x^{2} + 4) \cdot \frac{1}{x^{2} + 4}

\frac{\frac{d}{d x} (x^{2} + 4) \cdot 1}{x^{2} + 4}

\frac{\frac{d}{d x} (x^{2} + 4)}{x^{2} + 4}

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (x^{2} + 4)

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

\frac{\frac{d}{d x} (x^{2}) + 0}{x^{2} + 4}

ⓘ Adding 0 changes nothing:

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

\frac{d}{d x} (x^{2})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{2})

2 x

\frac{2 x}{x^{2} + 4}

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

ⓘ The derivative of

\frac{1}{2} \cdot \ln (x^{2} + 4)

\frac{x}{x^{2} + 4}

, so

\frac{1}{2} \cdot \ln (x^{2} + 4)

is an antiderivative of

\frac{x}{x^{2} + 4}

\int \frac{x}{x^{2} + 4} d x

\frac{1}{2} \cdot \ln (x^{2} + 4)

\frac{1}{2} \cdot \ln (x^{2} + 4)

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot \ln (x^{2} + 4)

\frac{1 \cdot \ln (x^{2} + 4)}{2}

\frac{\ln (x^{2} + 4)}{2}

▷

\frac{\ln (x^{2} + 4)}{2}

3) Let's evaluate for the interval

[-1, 3]

\frac{\ln (3^{2} + 4)}{2} - \frac{\ln ({(-1)}^{2} + 4)}{2}

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

3^{2}

3 \cdot 3

9

{(-1)}^{2}

(-1) \cdot (-1)

1

\frac{\ln (9 + 4)}{2} - \frac{\ln (1 + 4)}{2}

ⓘ Start from

9

and move forward

4

on the number line:

9 + 4

13

■Start from

1

and move forward

4

on the number line:

1 + 4

5

\frac{\ln (13)}{2} - \frac{\ln (5)}{2}

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{\ln (13)}{2} - \frac{\ln (5)}{2}

\frac{\ln (13) - \ln (5)}{2}

\frac{\ln (13) - \ln (5)}{2}

ⓘ For strictly positive arguments, a sum of logarithms is the logarithm of the product, a difference that of the quotient, ln(a) + ln(b) = ln(a*b):

\ln (13) - \ln (5)

\ln (\frac{13}{5})

▶Integral:

\frac{\ln (\frac{13}{5})}{2}

4) The numeric value of:

\frac{\ln (\frac{13}{5})}{2}

equals:

\frac{\ln (2.6)}{2}

\frac{\approx 0.9555}{2}

▷

\approx 0.4778