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

The integral from 1 to 4 of (x^2)/(x^2+4) is -2arctan(2)+2arctan(1/2)+3

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

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

- 2 arctan (2) + 2 arctan (\frac{1}{2}) + 3

Decimal

1.71299 \dots

Solution steps

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

\frac{x ^{2}}{x ^{2} + 4} = - \frac{4}{x ^{2} + 4} + 1

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

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot \int_{\frac{1}{2}}^{2} - \frac{1}{u ^{2} + 1} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{\frac{1}{2}}^{2} \frac{1}{u ^{2} + 1} d u = arctan (2) - arctan (\frac{1}{2})

\int_{\frac{1}{2}}^{2} 1 d u = \frac{3}{2}

= 2 (- (arctan (2) - arctan (\frac{1}{2})) + \frac{3}{2})

Simplify

2 (- (arctan (2) - arctan (\frac{1}{2})) + \frac{3}{2}) : - 2 arctan (2) + 2 arctan (\frac{1}{2}) + 3

= - 2 arctan (2) + 2 arctan (\frac{1}{2}) + 3

What is the integral from 1 to 4 of (x^2)/(x^2+4) ?
The integral from 1 to 4 of (x^2)/(x^2+4) is -2arctan(2)+2arctan(1/2)+3