# I don't get my mistake on how to solve 2int (1)/(x^2-x+1)dx, can you help me?

## I got this integral: Now at the step 2 we take out $\frac{3}{4}$, invert it in $\frac{4}{3}$ and multiply with $2$. But why we don't do the same thing at the step 4? So: $\frac{8}{3} \int \frac{1}{\frac{1}{3} {\left(2 x - 1\right)}^{2} + 1} \mathrm{dx}$ $3 \cdot \frac{8}{3} \int \frac{1}{{\left(2 x - 1\right)}^{2} + 1} \mathrm{dx}$ $8 \int \frac{1}{{\left(2 x - 1\right)}^{2} + 1} \mathrm{dx}$ and so on?

Jan 15, 2018

Because the denominator would be ${\left(2 x - 1\right)}^{2} + 3$ and not ${\left(2 x - 1\right)}^{2} + 1$

#### Explanation:

In order to use $\int \frac{1}{{t}^{2} + 1} \mathrm{dt} = {\tan}^{- 1} t + C$ we need the denominator to be of the form ${t}^{2} + 1$.

Attempting to eliminate the factor $\frac{1}{3}$ from $\frac{1}{3} {\left(2 x - 1\right)}^{2} + 1$ by multiplying by $3$ would result in ${\left(2 x - 1\right)}^{2} + 3$, which is not what we want.

So instead, we use $\frac{1}{3} {\left(2 x - 1\right)}^{2} = {\left(\frac{2 x - 1}{\sqrt{3}}\right)}^{2}$

Jan 15, 2018

There is a small arithmetic error here:

$\frac{8}{3} \int \frac{1}{\frac{1}{3} {\left(2 x - 1\right)}^{2} + 1} \mathrm{dx} \textcolor{red}{\ne} 3 \cdot \frac{8}{3} \int \frac{1}{{\left(2 x - 1\right)}^{2} + 1} \mathrm{dx}$

The correct working would be:

$\frac{8}{3} \int \frac{1}{\frac{1}{3} {\left(2 x - 1\right)}^{2} + 1} \mathrm{dx} = 3 \cdot \frac{8}{3} \int \frac{1}{{\left(2 x - 1\right)}^{2} \textcolor{red}{+ 3}} \mathrm{dx}$

A similar error is also made in going from step 3 to step 4.

So to answer you question, to factor that $\frac{1}{3}$ from the bottom would require multiplying that $1$ by 3, meaning that the direct integration would no longer work.

- A small recommendation:

There is a standard integral that:

$\int \frac{1}{{u}^{2} + {a}^{2}} \mathrm{du} = \frac{1}{a} \arctan \left(\frac{u}{a}\right) + C$

Given the substitution:
$u = x - \frac{1}{2}$
you could use this to evaluate the integral from directly from step 1.