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

The integral from 0 to x of 1/(t^2+t+1) is 2/(sqrt(3))arctan((2x+1)/(sqrt(3)))-pi/(3sqrt(3))

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

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

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

Solution steps

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

Complete the square

t^{2} + t + 1 : (t + \frac{1}{2})^{2} + \frac{3}{4}

= \int_{0}^{x} \frac{1}{( t + \frac{1}{2} ) ^{2} + \frac{3}{4}} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= 4 \cdot \int_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})} \frac{1}{2 3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 3} \cdot \int_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{2 3} [arctan (v)]_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})}

Simplify

4 \cdot \frac{1}{2 3} [arctan (v)]_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})} : \frac{2}{3} [arctan (v)]_{\frac{1}{3}}^{\frac{2 x + 1}{3}}

= \frac{2}{3} [arctan (v)]_{\frac{1}{3}}^{\frac{2 x + 1}{3}}

Compute the boundaries:

arctan (\frac{2 x + 1}{3}) - \frac{π}{6}

= \frac{2}{3} (arctan (\frac{2 x + 1}{3}) - \frac{π}{6})

Simplify

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

x \to 0 lim (sec (2 x))

\frac{1}{s - 4}

\int \frac{( 7 - 4 x ^{3} + 6 x ^{6} )}{x ^{6}} d x

\frac{d}{d x} (x^{5} (2 x - 1)^{7})

\frac{d}{d x} (\frac{x + 7}{x ^{3} + x - 6})

What is the integral from 0 to x of 1/(t^2+t+1) ?
The integral from 0 to x of 1/(t^2+t+1) is 2/(sqrt(3))arctan((2x+1)/(sqrt(3)))-pi/(3sqrt(3))