Evaluate the integral:


Evaluate the integral:

$\int \sqrt{x^{2}-2 x} d x$


Key points to solve the problem:

- Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have $\int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x)\left(\int g(x) d x\right) d x$

- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:

$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$

$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$

$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

Let, $I=\int \sqrt{x^{2}-2 x} d x$

We have:

$I=\int \sqrt{x^{2}-2 x} d x=\int \sqrt{x^{2}-2(1) x+1^{2}-1^{2}} d x$

Using $a^{2}-2 a b+b^{2}=(a-b)^{2}$

$I=\int \sqrt{(x-1)^{2}-1^{2}} d x$

As I match with the form:

$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$

$\therefore I$ $=\frac{x-1}{2} \sqrt{(x-1)^{2}-1}-\frac{1}{2} \log \left|x-1+\sqrt{(x-1)^{2}-1}\right|+C$

$\Rightarrow I$ $=\frac{x-1}{2} \sqrt{x^{2}-2 x}-\frac{1}{2} \log \left|x-1+\sqrt{x^{2}-2 x}\right|+C$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now