int sqrt{x^2 - 8x + 7} , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We need to evaluate the integral $I = \int \sqrt{x^2 - 8x + 7} \, dx$. First,complete the square for the quadratic expression: $x^2 - 8x + 7 = (x^

Vedclass · Accepted Answer

$\frac{1}{2}(x - 4)\sqrt{x^2 - 8x + 7} - \frac{9}{2}\log |x - 4 + \sqrt{x^2 - 8x + 7}| + c$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int \sqrt{x^2 - 8x + 7} \, dx$. First,complete the square for the quadratic expression: $x^2 - 8x + 7 = (x^2 - 8x + 16) - 16 + 7 = (x - 4)^2 - 9 = (x - 4)^2 - 3^2$. Now the integral becomes $I = \int \sqrt{(x - 4)^2 - 3^2} \, dx$. Using the standard formula $\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2}\sqrt{t^2 - a^2} - \frac{a^2}{2}\log |t + \sqrt{t^2 - a^2}| + c$,where $t = x - 4$ and $a = 3$: $I = \frac{x - 4}{2}\sqrt{(x - 4)^2 - 3^2} - \frac{3^2}{2}\log |(x - 4) + \sqrt{(x - 4)^2 - 3^2}| + c$. Simplifying this,we get $I = \frac{1}{2}(x - 4)\sqrt{x^2 - 8x + 7} - \frac{9}{2}\log |x - 4 + \sqrt{x^2 - 8x + 7}| + c$. Thus,the correct option is $C$.

$\int \sqrt{x^2 - 8x + 7} \, dx = $

Similar Questions

$\int \frac{x - 1}{(x + 1) \sqrt{x(x^2 + x + 1)}} dx =$

$\int \sqrt{\frac{1+x}{1-x}} \, dx = $

The value of $I = \int_{0}^{\frac{\pi}{4}} \tan^{n+1} x \, dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \tan^{n-1} \left( \frac{x}{2} \right) \, dx$ is

If $\int \operatorname{cosec}^5 x \, dx = \alpha \cot x \operatorname{cosec} x \left(\operatorname{cosec}^2 x + \frac{3}{2}\right) + \beta \log_e \left|\tan \frac{x}{2}\right| + C$,where $\alpha, \beta \in R$ and $C$ is the constant of integration,then the value of $8(\alpha + \beta)$ is equal to:

$\int {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } \;dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module