Best solutions to - "Сalculate the integral. ∫03dxx2+3"

permaneceerc

Answered question

2021-10-18

Сalculate the integral.

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

SchepperJ

Skilled2021-10-19Added 96 answers

Step 1
Given integral:

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

Step 2
Now,
Apply Integral Substitution:

x = \sqrt{3} u

d x = \sqrt{3} d u

And:
At

x = 0 \Rightarrow u = 0

x = 3 \Rightarrow u = \sqrt{3}

That is:

\int_{0}^{3} \frac{1}{x^{2} + 3} d x = \int_{0}^{\sqrt{3}} \frac{1}{\sqrt{3} (u^{2} + 1)} d u

Take the constant out:

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

= \frac{1}{\sqrt{3}} \cdot \int_{0}^{\sqrt{3}} \frac{1}{u^{2} + 1} d u

Use the common integral:

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

= \frac{1}{\sqrt{3}} {[\arctan (u)]}^{\sqrt{3}}

= \frac{1}{\sqrt{3}} \cdot [\frac{π}{3} - 0]

= \frac{π}{3 \sqrt{3}}

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