Question

$\int \frac{6x-8}{3x^2-2x+3}dx$

• A. In
• B. In
• C. In
• D. In

Answer 1

The correct option is A In

To solve such integrals we should always write linear expression in terms of the derivative of the quadratic expression.
So, 6x - 8 should be written in terms of the derivative of $3x^2-2x+3$
$\frac{d}{dx}(3x^2-2x+3)=6x-2$
So we write 6x - 8 = 6x - 2 - 6
So, the integral can be written as -
$\int \frac{(6x-2)-6}{3x^2-2x+3dx}$
Or $\int \frac{6x-2}{3x^2-2x+3_{dx}}-\int \frac{6}{3x^2-2x+3}${dx}
Let’s say $I_1=\int \frac{6x-2}{3x^2-2x+3}dx$
$I_2=\int \frac{6}{3x^2-2x+3dx}$
And solve each of these integrals separately.
$I_1=\int \frac{6x-2}{3x^2-2x+3_{dx}}$
Now to solve this, we’ll use the substitution method. We know that the numerator is nothing but the derivative of the denominator.
Let’s substitute $3x^2-2x+3$ =t
Then (6x - 2) dx = dt
So, the integral becomes -
$I_1=\int \frac{1}{t_{dt}}{I}_1=In\left|t\right|+C_1$
Let’s put the value of t in terms of x, which is $3x^2-2x+3$
$I_1=In|3x^2-2x+3|+C_1$
Now, let’s solve I_2
$I_2=\int \frac{6}{3x^2-2x+3}$dx
In this we can convert the given quadratic to a perfect square and we’ll use the corresponding formula
$I_2=\frac{6}{\int 3(x^2-\frac{2x}{3}+1)}dx$
$Or{I}_2\frac{2}{\int (x^2-\frac{2x}{3}+1)}dx$
$I_2=\frac{2}{\int (x^2-\frac{2x}{3}+\frac{1}{9}+\frac{8}{9})}$dx
$I_2=\frac{2}{\int {(x-\frac{1}{3})}^2+\frac{8}{9}}$dx
We can see that this is in $\int \frac{1}{x^2+a^2}dx$form.
So, we’ll use the corresponding formula .
$I_2=\frac{2}{\int {(x-\frac{1}{3})}^2}+\frac{8}{9}dx$
$I_2=2.\frac{\frac{1}{2^{\sqrt{2}}}}{3}ta{n}^{-}1\frac{x-1/3}{\left(2\sqrt{2}/3\right)}$
$Or{I}_2=\frac{3}{\sqrt{2}}{\tan }^{-1}\left(\frac{3x-1}{2\sqrt{2}}\right)+C_2$
So, the final answer would be =$I_1-I_2$
$=In|3x^2-2x+3|-\frac{3}{\sqrt{2}}{\tan }^{-1}\left(\frac{3x-1}{2\sqrt{2}}\right)+C$
Where $C,C_1,C_2$ are arbitrary constants and $C=C_1-C_2.$