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

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

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

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

\frac{2}{3} (arctan (\frac{2 x + 1}{3}) - arctan (33))

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{\frac{9}{2}}^{x + \frac{1}{2}} \frac{4}{4 u ^{2} + 3} d u

= \int_{\frac{9}{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{9}{2}}^{x + \frac{1}{2}} \frac{1}{4 u ^{2} + 3} d u

Apply Integral Substitution

= 4 \cdot \int_{33}^{\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_{33}^{\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)]_{33}^{\frac{2}{3} (x + \frac{1}{2})}

Simplify

4 \cdot \frac{1}{2 3} [arctan (v)]_{33}^{\frac{2}{3} (x + \frac{1}{2})} : \frac{2}{3} [arctan (v)]_{33}^{\frac{2 x + 1}{3}}

= \frac{2}{3} [arctan (v)]_{33}^{\frac{2 x + 1}{3}}

Compute the boundaries:

arctan (\frac{2 x + 1}{3}) - arctan (33)

= \frac{2}{3} (arctan (\frac{2 x + 1}{3}) - arctan (33))

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