If \int \frac(\sqrt1+x^2+x)^10(\sqrt1+x^2-x)^9 ~d x=\frac1~m((\sqrt1+x^2+x)^n(n \sqrt1+x^2-x))+C where C is the constant of integration and m, n \in N , then m+n is equal to \underline\hspace2cm .

Indefinite Integrals PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q46 with Solution

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Correct answer: 379

Simplify the integrand

Let $u=\sqrt{1+x^2}+x.$ Then $\left(\sqrt{1+x^2}+x\right)\left(\sqrt{1+x^2}-x\right)=1,$ so $\sqrt{1+x^2}-x=\frac{1}{u}.$

Hence the integrand becomes $\frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9}=u^{10}\cdot u^9=u^{19}.$

So we need to evaluate $\int u^{19}\,dx.$

Find a useful derivative

We know $u=\sqrt{1+x^2}+x.$ Differentiate: $\frac{du}{dx}=\frac{x}{\sqrt{1+x^2}}+1=\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}=\frac{u}{\sqrt{1+x^2}}.$ Thus,

\quad\Rightarrow\quad dx=\frac{\sqrt{1+x^2}}{u}\,du.$$ Therefore, $$\int u^{19}dx=\int u^{19}\cdot \frac{\sqrt{1+x^2}}{u}\,du=\int u^{18}\sqrt{1+x^2}\,du.$$ Now express $\sqrt{1+x^2}$ in terms of $u$. Since $$u=\sqrt{1+x^2}+x, \qquad \frac1u=\sqrt{1+x^2}-x,$$ adding, $$u+\frac1u=2\sqrt{1+x^2}.$$ So $$\sqrt{1+x^2}=\frac12\left(u+\frac1u\right).$$ Hence $$\int u^{18}\sqrt{1+x^2}\,du =\frac12\int u^{18}\left(u+\frac1u\right)du =\frac12\int (u^{19}+u^{17})du.$$ This gives $$\frac12\left(\frac{u^{20}}{20}+\frac{u^{18}}{18}\right)+C =\frac{u^{20}}{40}+\frac{u^{18}}{36}+C.$$ Taking LCM $360$, $$\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9}dx =\frac1{360}\left(9u^{20}+10u^{18}\right)+C =\frac1{360}u^{18}(9u^2+10)+C.$$ --- 3. **Match with the given form** We are given that $$\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9}dx =\frac1m\left((\sqrt{1+x^2}+x)^n\,(n\sqrt{1+x^2}-x)\right)+C.$$ Now put back $u=\sqrt{1+x^2}+x$. We claim $n=19$. Check: $$u^{19}(19\sqrt{1+x^2}-x).$$ Using $$x=\frac12\left(u-\frac1u\right), \qquad \sqrt{1+x^2}=\frac12\left(u+\frac1u\right),$$ we get $$19\sqrt{1+x^2}-x =\frac{19}{2}\left(u+\frac1u\right)-\frac12\left(u-\frac1u\right) =\frac12\left(18u+20\frac1u\right) =9u+\frac{10}{u}.$$ Therefore, $$u^{19}(19\sqrt{1+x^2}-x)=9u^{20}+10u^{18}.$$ So indeed, $$\int \cdots dx=\frac1{360}\,u^{19}(19\sqrt{1+x^2}-x)+C.$$ Thus, $$m=360,\qquad n=19.$$ Therefore, $$m+n=360+19=379.$$ --- 4. **Comparison with stored answer** Stored correct answer = $379$. Our derived answer also is $379$, so it agrees.

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View written solutionFree

Correct answer: 379

Simplify the integrand

Let $u=\sqrt{1+x^2}+x.$ Then $\left(\sqrt{1+x^2}+x\right)\left(\sqrt{1+x^2}-x\right)=1,$ so $\sqrt{1+x^2}-x=\frac{1}{u}.$

Hence the integrand becomes $\frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9}=u^{10}\cdot u^9=u^{19}.$

So we need to evaluate $\int u^{19}\,dx.$

Find a useful derivative

We know $u=\sqrt{1+x^2}+x.$ Differentiate: $\frac{du}{dx}=\frac{x}{\sqrt{1+x^2}}+1=\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}=\frac{u}{\sqrt{1+x^2}}.$ Thus,

\quad\Rightarrow\quad dx=\frac{\sqrt{1+x^2}}{u}\,du.$$ Therefore, $$\int u^{19}dx=\int u^{19}\cdot \frac{\sqrt{1+x^2}}{u}\,du=\int u^{18}\sqrt{1+x^2}\,du.$$ Now express $\sqrt{1+x^2}$ in terms of $u$. Since $$u=\sqrt{1+x^2}+x, \qquad \frac1u=\sqrt{1+x^2}-x,$$ adding, $$u+\frac1u=2\sqrt{1+x^2}.$$ So $$\sqrt{1+x^2}=\frac12\left(u+\frac1u\right).$$ Hence $$\int u^{18}\sqrt{1+x^2}\,du =\frac12\int u^{18}\left(u+\frac1u\right)du =\frac12\int (u^{19}+u^{17})du.$$ This gives $$\frac12\left(\frac{u^{20}}{20}+\frac{u^{18}}{18}\right)+C =\frac{u^{20}}{40}+\frac{u^{18}}{36}+C.$$ Taking LCM $360$, $$\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9}dx =\frac1{360}\left(9u^{20}+10u^{18}\right)+C =\frac1{360}u^{18}(9u^2+10)+C.$$ --- 3. **Match with the given form** We are given that $$\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9}dx =\frac1m\left((\sqrt{1+x^2}+x)^n\,(n\sqrt{1+x^2}-x)\right)+C.$$ Now put back $u=\sqrt{1+x^2}+x$. We claim $n=19$. Check: $$u^{19}(19\sqrt{1+x^2}-x).$$ Using $$x=\frac12\left(u-\frac1u\right), \qquad \sqrt{1+x^2}=\frac12\left(u+\frac1u\right),$$ we get $$19\sqrt{1+x^2}-x =\frac{19}{2}\left(u+\frac1u\right)-\frac12\left(u-\frac1u\right) =\frac12\left(18u+20\frac1u\right) =9u+\frac{10}{u}.$$ Therefore, $$u^{19}(19\sqrt{1+x^2}-x)=9u^{20}+10u^{18}.$$ So indeed, $$\int \cdots dx=\frac1{360}\,u^{19}(19\sqrt{1+x^2}-x)+C.$$ Thus, $$m=360,\qquad n=19.$$ Therefore, $$m+n=360+19=379.$$ --- 4. **Comparison with stored answer** Stored correct answer = $379$. Our derived answer also is $379$, so it agrees.

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