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

The integral from 1 to 4 of 3/(1+4x^2) is 3/2 (arctan(8)-arctan(2))

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integral from 1 to 4 of 3/(1+4x^2)

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

\frac{3}{2} (arctan (8) - arctan (2))

Decimal

0.50893 \dots

Solution steps

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

Apply Integral Substitution

= 3 \cdot \int_{2}^{8} \frac{1}{2 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{2}^{8} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} [arctan (u)]_{2}^{8}

Simplify

3 \cdot \frac{1}{2} [arctan (u)]_{2}^{8} : \frac{3}{2} [arctan (u)]_{2}^{8}

= \frac{3}{2} [arctan (u)]_{2}^{8}

Compute the boundaries:

arctan (8) - arctan (2)

= \frac{3}{2} (arctan (8) - arctan (2))

\int_{- \infty}^{1} e^{- x} d x

\int_{1}^{2} \frac{8}{x} d x

\int_{- 1}^{0} 18 x^{2} (3 x^{3} + 3)^{2} d x

\int_{4}^{6} x^{2} - 2 x - 8 d x

\int_{4}^{19} \frac{18}{x ^{3}} d x

What is the integral from 1 to 4 of 3/(1+4x^2) ?
The integral from 1 to 4 of 3/(1+4x^2) is 3/2 (arctan(8)-arctan(2))