\(\rm \int \sqrt{4x-3}\;dx\) is equal to?

Question

Question

Answer 1

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1} +c\)

Calculation:

I = \(\rm \int \sqrt{4x-3}\;dx\)

Let 4x - 3 = t²

Differenating with respect to x, we get

⇒ 4dx = 2tdt

⇒ dx = \(\rm \frac t 2\)dt

Now,

I = \(\rm \int \sqrt{t^2}\; \times \frac t 2 dt\)

= \(\frac12 \rm \int t^2 \;dt\)

= \(\rm \frac {t^3}{6} + c\)

= \(\rm \frac {(4x-3)^{3/2}}{6} + c\)