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

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

Question

(D) To solve the integral $\int \sqrt{x^2-8 x+7} \, dx$,we first complete the square for the quadratic expression inside the square root:$x^2 - 8x + 7

Vedclass · Accepted Answer

$\frac{x-4}{2} \sqrt{x^2-8 x+7} - \frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|$

Vedclass · Answer

(D) To solve the integral $\int \sqrt{x^2-8 x+7} \, dx$,we first complete the square for the quadratic expression inside the square root:$x^2 - 8x + 7 = (x^2 - 8x + 16) - 16 + 7 = (x-4)^2 - 9 = (x-4)^2 - 3^2$.Now the integral becomes $\int \sqrt{(x-4)^2 - 3^2} \, dx$.Using the standard formula $\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2} \sqrt{t^2 - a^2} - \frac{a^2}{2} \log \left|t + \sqrt{t^2 - a^2}\right| + C$,where $t = x-4$ and $a = 3$:$= \frac{x-4}{2} \sqrt{(x-4)^2 - 3^2} - \frac{3^2}{2} \log \left|(x-4) + \sqrt{(x-4)^2 - 3^2}\right| + C$$= \frac{x-4}{2} \sqrt{x^2-8x+7} - \frac{9}{2} \log \left|x-4 + \sqrt{x^2-8x+7}\right| + C$.Comparing this with the given options,the correct option is $D$.

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

Similar Questions

$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$ is equal to :

If $\int e^{3x} \cos 4x \,dx = e^{3x} (A \sin 4x + B \cos 4x) + c$,then:

$\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x=$

Let for $f(x)=7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$,$I_1 = \int_0^{\pi/4} f(x) \, dx$ and $I_2 = \int_0^{\pi/4} x f(x) \, dx$. Then $7 I_1 + 12 I_2$ is equal to:

$\int \frac{d x}{(x+1) \sqrt{x^2+4}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module