What is the integral from 0 to x of 6/(1+6x^2) ?

The integral from 0 to x of 6/(1+6x^2) is sqrt(6)arctan(sqrt(6)x)

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integral from 0 to x of 6/(1+6x^2)

\int_{0}^{x} \frac{6}{1 + 6 x ^{2}} d x

6 arctan (6 x)

Solution steps

\int_{0}^{x} \frac{6}{1 + 6 x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{x} \frac{1}{1 + 6 x ^{2}} d x

Apply Integral Substitution

= 6 \cdot \int_{0}^{6 x} \frac{1}{6 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{0}^{6 x} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} [arctan (u)]_{0}^{6 x}

Simplify

6 \cdot \frac{1}{6} [arctan (u)]_{0}^{6 x} : 6 [arctan (u)]_{0}^{6 x}

= 6 [arctan (u)]_{0}^{6 x}

Compute the boundaries:

arctan (6 x)

= 6 arctan (6 x)

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What is the integral from 0 to x of 6/(1+6x^2) ?
The integral from 0 to x of 6/(1+6x^2) is sqrt(6)arctan(sqrt(6)x)