int sqrt{x^{2}-8 x+7} , dx is equal to | vedclass

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

Question

(A) Let $I = \int \sqrt{x^{2}-8 x+7} \, dx$. Completing the square,we have $x^{2}-8x+7 = (x-4)^{2}-16+7 = (x-4)^{2}-3^{2}$. Thus,$I = \int \sqrt{(x-4)

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \sqrt{x^{2}-8 x+7} \, dx$. Completing the square,we have $x^{2}-8x+7 = (x-4)^{2}-16+7 = (x-4)^{2}-3^{2}$. Thus,$I = \int \sqrt{(x-4)^{2}-3^{2}} \, dx$. Using the standard formula $\int \sqrt{x^{2}-a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}-a^{2}} - \frac{a^{2}}{2} \log |x + \sqrt{x^{2}-a^{2}}| + C$,where $x$ is replaced by $(x-4)$ 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 the expression,we get $I = \frac{1}{2}(x-4) \sqrt{x^{2}-8x+7} - \frac{9}{2} \log |x-4+\sqrt{x^{2}-8x+7}| + C$. Therefore,the correct option is $A$.

$\int \sqrt{x^{2}-8 x+7} \, dx$ is equal to

Similar Questions

$\int \frac{d x}{12 \cos x+5 \sin x}=$

The value of $\int_{\frac{1}{3}}^1 (x - x^3)^{\frac{1}{3}} dx$ is

$\int \frac{x}{\sqrt{x^2-2x+5}} dx=$

If $\int(x^6+x^4+x^2) \sqrt{2x^4+3x^2+6} dx = f(x) + c$,then $f(3) =$

If $\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx = \sin^{-1}(f(x)) + C$,(where $C$ is the constant of integration),then $f(2)$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module