Question

$\int \frac{10x-5}{\sqrt{x^2-x+3}dx}$ will be -

• A. 5$\sqrt{x^2-x+3}+C$
• B. 3$\sqrt{x^2-x+3}+C$
• C. 10$\sqrt{x^2-x+3}+C$
• D. $\text{None of these}$

Answer 1

The correct option is C 10$\sqrt{x^2-x+3}+C$

Here also we have one linear expression and one quadratic. We will express the linear expression in terms of derivative of quadratic expression.
$\frac{d}{dx}(x^2-x+3)=2x-1$
So, 10x - 5=5(2x -1)
And the integral could be written as -
5$\int \frac{2x-1}{\sqrt{x^2-x+3}}dx$
Let’s apply the substitution method
Let$x^2-x+3=t^2$
So, (2x - 1 ) dx = 2t .dt
And the integral would be -
$5\int \frac{2t}{t}dtOr10\int 1.dt=10\left(t{+}^{C_1}\right)10t+C$
Let’s substitute the value of “t” which is $\sqrt{x^2-x+3}$
So, the final answer would be = 10 $\sqrt{x^2-x+3}$ +C