Animated Solution for Mathematics - Indefinite Integration: If $\int {\sin }^{-1}\left(\sqrt{\frac{x}{1+x}}\right)dx=A(x){\tan }^{-1}(\sqrt{x})+B(x)+C$ where $C$ is a constant of integration, then the ordered pair $(A(x),B(x))$ can be

Welcome, future engineer. When you first look at an integral like $\int {\sin }^{-1}\left(\sqrt{\frac{x}{1+x}}\right)dx$, it is natural to feel a moment of hesitation. It looks like a tangled mess of radicals and inverse functions.

But in the world of JEE Advanced, complexity is often just a mask for a hidden, elegant simplicity. Our job is to peel back the layers.

The Power of Substitution

The first thing that screams for attention is the term $\sqrt{\frac{x}{1+x}}$. That radical is the source of all our trouble. We need to liberate $x$ from it.

By choosing the substitution $x=t^2$, we immediately transform the radical into $\sqrt{\frac{t^2}{1+t^2}}$, which simplifies beautifully to $\frac{t}{\sqrt{1+t^2}}$.

Remember, whenever you substitute, you must also transform the differential. Since $x=t^2$, we have $dx=2tdt$.

Now, our integral is no longer a terrifying expression in $x$; it is a manageable expression in $t$. We have successfully navigated the first hurdle.

The Geometric Insight

Now we face ${\sin }^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$. Many students try to differentiate this using the chain rule, which leads to a nightmare of algebra. Stop! Pause! Look at the expression geometrically.

Imagine a right-angled triangle. If we define an angle $\varphi$ such that $\sin \varphi =\frac{t}{\sqrt{1+t^2}}$, then the perpendicular side is $t$ and the hypotenuse is $\sqrt{1+t^2}$.

By the Pythagorean theorem, the base must be:

If the perpendicular is $t$ and the base is $1$, then $\tan \varphi =\frac{t}{1}=t$. This means $\varphi ={\tan }^{-1}(t)$.

Just like that, the complex inverse sine function collapses into a simple ${\tan }^{-1}(t)$. This is the kind of insight that separates the top rankers from the rest—the ability to see the geometry hidden within the algebra.

The IBP Dance

With our substitutions, the integral becomes $I=\int ({\tan }^{-1}t)\cdot (2t)dt$. We have a product of an inverse trigonometric function and an algebraic function. This is the classic playground for Integration by Parts (IBP).

Following the ILATE rule, we choose $u={\tan }^{-1}t$ and $dv=2tdt$.

Differentiating $u$, we get $du=\frac{1}{1+t^2}dt$. Integrating $dv$, we get $v=t^2$. Applying the IBP formula $\int udv=uv-\int vdu$, we arrive at:

The Final Algebraic Cleanup

We are almost at the finish line. The remaining integral is $\int \frac{t^2}{1+t^2}dt$. This is a classic JEE trap.

When the degree of the numerator equals the degree of the denominator, we use the 'add and subtract' trick. We rewrite the numerator $t^2$ as $t^2+1-1$.

This allows us to split the integral:

This is trivial to solve! It becomes $t-{\tan }^{-1}t$.

Bringing it Home

Finally, we combine everything. Substituting back into our IBP result, we get:

Distributing the negative sign and grouping the ${\tan }^{-1}t$ terms, we get $(t^2+1){\tan }^{-1}t-t+C$.

Since $t=\sqrt{x}$, we replace $t$ to get the final form: $(x+1){\tan }^{-1}(\sqrt{x})-\sqrt{x}+C$.

Comparing this to the required form $A(x){\tan }^{-1}(\sqrt{x})+B(x)+C$, we identify $A(x)=x+1$ and $B(x)=-\sqrt{x}$. You have conquered the beast. Remember, every complex integral is just a series of small, logical steps waiting to be taken.