\(\rm \displaystyle\int_0^{16} \frac{x^{\frac{1}{4}}}{{1+ \sqrt{x}}}dx\) का मान क्या है?

गणना:

माना, \(I=\rm \displaystyle\int_0^{16} \frac{x^{\frac{1}{4}}}{{1+ \sqrt{x}}}dx\)

t = \(\rm x^\frac{1}{4}\) ⇒ x = t⁴ ⇒ dx = 4t³dt प्रतिस्थापित करने पर,

जब, x = 0 ⇒ t = 0

और x = 16 ⇒ t = 2

∴\(I=\rm \displaystyle \int_{0}^2\frac{t}{1+t^2}\times 4t^3dt\)

= \(\rm \displaystyle 4\int_{0}^2\frac{t^4}{1+t^2}dt\)

= \(\rm \displaystyle 4\int_{0}^2\left( \frac{1}{1+t^2}+t^2-1\right )dt\)

= \(\rm 4\left[tan^{-1}t+\frac{t^3}{3}-t\right]_{0}^{2}\)

= \(\rm 4(\left[tan^{-1}(2)+\frac{2^3}{3}-2\right]-0)\)

= \(\rm 4\left[ tan^{-1}(2)+\frac{2}{3}\right ]\)

\(=\frac{8}{3}+4\tan^{-1}2\)

∴ \(\rm \displaystyle\int_0^{16} \frac{x^{\frac{1}{4}}}{{1+ \sqrt{x}}}dx\)  \(=\frac{8}{3}+4\tan^{-1}2\)