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

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

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

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

3 (arctan (a) - arctan (- a))

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= 3 [arctan (x)]_{- a}^{a}

Compute the boundaries:

arctan (a) - arctan (- a)

= 3 (arctan (a) - arctan (- a))

\int_{0}^{4} (6 t - 3) d t

\int_{2}^{3} 4 d x

\int_{- 1}^{1} 6 x d x

\int_{0}^{9} 5 t d t

\int_{0}^{1000} 0.03 q + 0.1 d q

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