JEE Main 2022 — Mathematics Calculus

Given,

$\displaystyle {\int }_{0}^{\sqrt{3}}\frac{15{x}^{3}}{\sqrt{1+{x}^{2}+\sqrt{{(1+{x}^{2})}^{3}}}}dx=\alpha \sqrt{2}+\beta \sqrt{3}$∫03​​1+x2+(1+x2)3​​15x3​dx=α2​+β3​,

Taking L.H.S we get $\displaystyle {\int }_{0}^{\sqrt{3}}\frac{15{x}^{3}}{\sqrt{1+{x}^{2}+\sqrt{{(1+{x}^{2})}^{3}}}}dx$∫03​​1+x2+(1+x2)3​​15x3​dx

Now put $1+{x}^{2}={t}^{2}$1+x2=t2

$$\Rightarrow 2xdx=2tdt$$⇒2xdx=2tdt

$$\Rightarrow xdt=tdt$$⇒xdt=tdt

$$\therefore {\int }_{1}^{2}\frac{15({t}^{2}-1)tdt}{\sqrt{{t}^{2}+{t}^{3}}}$$∴∫12​t2+t3​15(t2−1)tdt​

$$=15{\int }_{1}^{2}\frac{t({t}^{2}-1)}{t\sqrt{1+t}}dt$$=15∫12​t1+t​t(t2−1)​dt

Now put $1+t={u}^{2}$1+t=u2$\Rightarrow dt=2udu$⇒dt=2udu

So, the integral will be $\displaystyle 15{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{{({u}^{2}-1)}^{2}-1}{u}\times 2udu$15∫2​3​​u(u2−1)2−1​×2udu

$$=30{\int }_{\sqrt{2}}^{\sqrt{3}}({u}^{4}-2{u}^{2})du$$=30∫2​3​​(u4−2u2)du

$$=30{(\frac{{u}^{5}}{5}-\frac{2{u}^{3}}{3})}_{\sqrt{2}}^{\sqrt{3}}$$=30(5u5​−32u3​)2​3​​

$$=30[\frac{1}{5}({\sqrt{3}}^{5}-{\sqrt{2}}^{5})-\frac{2}{3}({\sqrt{3}}^{3}-{\sqrt{2}}^{3})]$$=30[51​(3​5−2​5)−32​(3​3−2​3)]

$$=30[\frac{1}{5}(9\sqrt{3}-4\sqrt{2})-\frac{2}{3}(3\sqrt{3}-2\sqrt{2})]$$=30[51​(93​−42​)−32​(33​−22​)]

$$=30[-\frac{1}{5}\times \sqrt{3}+\frac{8}{15}\sqrt{2}]$$=30[−51​×3​+158​2​]

$$=-6\sqrt{3}+16\sqrt{2}$$=−63​+162​

So, $\displaystyle {\int }_{0}^{\sqrt{4}}\frac{15{x}^{3}}{\sqrt{1+{x}^{2}+\sqrt{{(1+{x}^{2})}^{3}}}}dx=\alpha \sqrt{2}+\beta \sqrt{3}$∫04​​1+x2+(1+x2)3​​15x3​dx=α2​+β3​

$$\Rightarrow -6\sqrt{3}+16\sqrt{2}=\alpha \sqrt{2}+\beta \sqrt{3}$$⇒−63​+162​=α2​+β3​

Now on comparing we get, $\alpha =16,\beta =-6$α=16,β=−6

$$\therefore \alpha +\beta =10$$∴α+β=10