Evaluate the integral:

Question:

Evaluate the integral:

Solution:

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:

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

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

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

Let, $I=\int \sqrt{4 x^{2}-5} d x$

We have:

$I=\int \sqrt{4 x^{2}-5} d x=\int 2 \sqrt{x^{2}-\frac{5}{4}} d x$

$\Rightarrow I=2 \int \sqrt{x^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}} d x$

As I match with the form:

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

$\therefore I=2\left\{\frac{x}{2} \sqrt{x^{2}}-\left(\frac{\sqrt{5}}{2}\right)^{2}-\frac{\frac{5}{4}}{2} \log \left|x+\sqrt{x^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}}\right|\right\}$

$\Rightarrow I=x \sqrt{x^{2}-\frac{5}{4}}-\frac{5}{4} \log \left|x+\sqrt{x^{2}-\frac{5}{4}}\right|+C$

Leave a comment

Close

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