Question

$\int {\sqrt{2x^2-3x+1}}_{dx}$ will be equal to -

• A. $\sqrt{2}(\frac{(x-3/4)}{2}\sqrt{x^2-1/36}-\frac{1}{32}log|(x-3/4)+\sqrt{{(x-3/4)}^2-1/36}|)$ +C
• B. $\sqrt{2}(\frac{(x-3/4)}{2}\sqrt{{(x-3/4)}^2-1/16}-\frac{1}{32}log|(x-3/4)+\sqrt{{(x-3/4)}^2-1/16}|)$+C
• C. $\sqrt{2}(\frac{(x-3/4)}{2}\sqrt{x^2-1/36}-\frac{1}{32}log|(x-3/4)+\sqrt{{(x-3/4)}^2-1/36}|)$ +C
• D. $\sqrt{2}(\frac{(x-3/4)}{2}\sqrt{x^2-1/16}-\frac{1}{32}log|(x-3/4)+\sqrt{{(x-3/2)}^2-1/16}|)$ +C

Answer 1

The correct option is B $\sqrt{2}(\frac{(x-3/4)}{2}\sqrt{{(x-3/4)}^2-1/16}-\frac{1}{32}log|(x-3/4)+\sqrt{{(x-3/4)}^2-1/16}|)$+C

We’ll use the same approach here also. We’ll make the given expression inside the underroot sign a perfect square and then apply the corresponding formula
$\int \sqrt{2x^2-3x+1}dxOr\int \sqrt{2}\sqrt{x^2-3x/2+1/2}dxOr\int \sqrt{2}\sqrt{x^2-3x/2+9/16-9/16+1/2}dxOr\int \sqrt{2}\sqrt{{(x-3/4)}^2-1/16}dx$
We can see that this is$\int \sqrt{x^2-a^2}dx$ form and the corresponding formula is$(\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}log|x+\sqrt{x^2-a^2}|)$+ C
Here, instead of x we have $(x-\frac{3}{4})$ and value of “a” is $\frac{1}{4}$.
On putting the values we get,
$\sqrt{2}(\frac{(x-3/4)}{2}\sqrt{{(x-3/4)}^2-1/16}-\frac{1}{32}log|(x-3/4)+\sqrt{{(x-3/4)}^2-1/16}|)$+C