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

$\int \sqrt{x^2+x+1} \, dx \times \int \frac{1}{\sqrt{x^2+x+1}} \, dx$ is equal to

$\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x=A \cos x+B \log |f(x)|+c$ (where $c$ is a constant of integration). Then the values of $A, B$ and $f(x)$ are:

Find $\int \sqrt{x^{2}+2 x+5} \, dx$.

Given that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$. If $\int \frac{1}{x^4+3x^2+1} dx = a \cdot \tan^{-1}\left(\frac{b(x^2-1)}{x}\right) + c \cdot \tan^{-1}\left(\frac{d(x^2+1)}{x}\right) + k$,where $k$ is a constant of integration,then $5(c+d+ab) = $

$\int \frac{x+\sin x}{1+\cos x} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module