If \int\limits_0^\sqrt3 \frac15 x^3\sqrt1+x^2+\sqrt(1+x^2)^3 ~d x=\alpha \sqrt2+\beta \sqrt3 , where \alpha, \beta are integers, then \alpha+\beta is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2022 · 28 Jul · Shift 1 · Q41 with Solution

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Correct answer: 10

We need to evaluate $I=\int_{0}^{\sqrt{3}} \frac{15x^3}{\sqrt{1+x^2+\sqrt{(1+x^2)^3}}}\,dx.$

Since $1+x^2\ge 0$ , we have $\sqrt{(1+x^2)^3}=(1+x^2)^{3/2}.$ So the denominator becomes $\sqrt{1+x^2+(1+x^2)^{3/2}}.$

Let $t=\sqrt{1+x^2}.$ Then $1+x^2=t^2, \qquad (1+x^2)^{3/2}=t^3.$ Hence denominator: $\sqrt{t^2+t^3}=\sqrt{t^2(1+t)}=t\sqrt{1+t}$ because $t\ge 0$ .

Thus $I=\int_0^{\sqrt3} \frac{15x^3}{t\sqrt{1+t}}\,dx.$

From $t=\sqrt{1+x^2} \implies t^2=1+x^2,$ we get $2t\,dt=2x\,dx \implies x\,dx=t\,dt.$ Also, $x^2=t^2-1 \implies x^3dx=x^2(xdx)=(t^2-1)(t\,dt).$

Therefore, $I=\int \frac{15(t^2-1)t}{t\sqrt{1+t}}\,dt=15\int \frac{t^2-1}{\sqrt{1+t}}\,dt.$

Now change limits:

So $I=15\int_1^2 \frac{t^2-1}{\sqrt{1+t}}\,dt.$

Let $u=1+t \implies t=u-1, \quad dt=du.$ When $t=1$ , $u=2$ ; when $t=2$ , $u=3$ .

Also, $t^2-1=(u-1)^2-1=u^2-2u.$ Thus $\frac{t^2-1}{\sqrt{1+t}}=\frac{u^2-2u}{\sqrt u}=u^{3/2}-2u^{1/2}.$

Hence, $I=15\int_2^3 \left(u^{3/2}-2u^{1/2}\right)du.$

$\int u^{3/2}du=\frac{2}{5}u^{5/2}, \qquad \int u^{1/2}du=\frac{2}{3}u^{3/2}.$ So

=15\left[\frac{2}{5}u^{5/2}-\frac{4}{3}u^{3/2}\right]_2^3.$$ Factor $u^{3/2}$: $$\frac{2}{5}u^{5/2}-\frac{4}{3}u^{3/2}=u^{3/2}\left(\frac{2}{5}u-\frac{4}{3}\right).$$ Now evaluate: At $u=3$, $$\frac{2}{5}3^{5/2}-\frac{4}{3}3^{3/2} =\frac{2}{5}(9\sqrt3)-\frac{4}{3}(3\sqrt3) =\frac{18}{5}\sqrt3-4\sqrt3 =-\frac{2}{5}\sqrt3.$$ At $u=2$, $$\frac{2}{5}2^{5/2}-\frac{4}{3}2^{3/2} =\frac{2}{5}(4\sqrt2)-\frac{4}{3}(2\sqrt2) =\frac{8}{5}\sqrt2-\frac{8}{3}\sqrt2 =-\frac{16}{15}\sqrt2.$$ Therefore, $$I=15\left(-\frac{2}{5}\sqrt3-\left(-\frac{16}{15}\sqrt2\right)\right) =15\left(-\frac{2}{5}\sqrt3+\frac{16}{15}\sqrt2\right).$$ Multiply by $15$: $$I=-6\sqrt3+16\sqrt2.$$ 5. **Identify $\alpha$ and $\beta$** Given $$I=\alpha\sqrt2+\beta\sqrt3,$$ we get $$\alpha=16, \qquad \beta=-6.$$ So $$\alpha+\beta=16+(-6)=10.$$ 6. **Comparison with stored answer** Derived answer is $10$, which matches the stored correct answer.