If ∫ (√(1+x²)+x)¹⁰ / (√(1+x²)-x)⁹ dx = 1/m ((√(1+x²)+x)ⁿ (n√(1+x²)-x)) + C where C is the constant o

Q: Why did we choose $t = \sqrt{1+x^2} + x$ instead of a standard trigonometric substitution like $x = \tan\theta$?

While trigonometric substitution is a valid path, it would lead to a very complex integral involving powers of secant and tangent. The conjugate substitution $t = \sqrt{1+x^2} + x$ is a specialized technique for expressions involving $\sqrt{1+x^2} \pm x$. It turns the entire expression into a simple power function of $t$, which is much faster and less prone to calculation errors.

Q: How did we know the differential $dx$ would simplify so nicely?

It is a beautiful property of this specific substitution. When you differentiate $x = \frac{1}{2}(t - t^{-1})$, the result is $\frac{1}{2}(1 + t^{-2})dt$, which is $\frac{t^2+1}{2t^2}dt$. This term perfectly complements the $t$ terms in the integral, leading to a clean polynomial.

Q: Why did we split the term $\frac{10}{t}$ into $\frac{9}{t} + \frac{1}{t}$ in the final steps?

This is a strategic manipulation to match the target form given in the problem. The target form requires a specific structure: $(\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x)$. By splitting the term, we create a factor of $9(t + 1/t)$, which corresponds to $9(2\sqrt{1+x^2}) = 18\sqrt{1+x^2}$, allowing us to combine it with the remaining $\sqrt{1+x^2}$ to get $19\sqrt{1+x^2}$.

Question

If $\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx = \frac{1}{m} ((\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x)) + C$ where $C$ is the constant of integration and $m, n \in \mathbb{N}$, then $m+n$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a towering, intimidating mountain of an integral:

$$I = \int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx$$

At first glance, it seems impossible. But in the world of JEE Advanced, every monster has a weakness. Our weakness here is the hidden symmetry of conjugate pairs.

### The Conjugate Insight

Look closely at the base of the numerator, $(\sqrt{1+x^2}+x)$, and the base of the denominator, $(\sqrt{1+x^2}-x)$. They are conjugates. If you multiply them together, the $x^2$ terms cancel out, leaving you with $1$.

This means the denominator is simply the reciprocal of the numerator's base. Specifically, since $(\sqrt{1+x^2}+x)(\sqrt{1+x^2}-x) = 1$, we have:

$$\frac{1}{(\sqrt{1+x^2}-x)^9} = (\sqrt{1+x^2}+x)^9$$

Thus, the integral simplifies to:

$$I = \int (\sqrt{1+x^2}+x)^{10} \cdot (\sqrt{1+x^2}+x)^9 dx = \int (\sqrt{1+x^2}+x)^{19} dx$$

### The Strategic Substitution

To solve this, we use the substitution $t = \sqrt{1+x^2} + x$. We know that $1/t = \sqrt{1+x^2} - x$.

If we add these two equations, we get $t + 1/t = 2\sqrt{1+x^2}$. If we subtract them, we get $t - 1/t = 2x$, which implies $x = \frac{1}{2}(t - 1/t)$.

Now, we find the differential $dx$. Differentiating $x$ with respect to $t$ gives:

$$dx = \frac{1}{2}\left(1 + \frac{1}{t^2}\right) dt = \frac{t^2+1}{2t^2} dt$$

### The Transformation

Substituting everything back into our integral, the expression becomes:

$$I = \int t^{19} \cdot \frac{t^2+1}{2t^2} dt$$

Simplifying this, we obtain:

$$I = \frac{1}{2} \int t^{17}(t^2+1) dt = \frac{1}{2} \int (t^{19} + t^{17}) dt$$

Applying the power rule, we integrate to find:

$$I = \frac{1}{2} \left[ \frac{t^{20}}{20} + \frac{t^{18}}{18} \right] + C = \frac{t^{20}}{40} + \frac{t^{18}}{36} + C$$

### The Final Alignment

We need to match the target form: $\frac{1}{m} ((\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x)) + C$. We factor out $\frac{t^{19}}{360}$ to get:

$$I = \frac{t^{19}}{360} \left[ 9t + \frac{10}{t} \right] + C$$

Splitting $\frac{10}{t}$ into $\frac{9}{t} + \frac{1}{t}$, we rewrite the expression as:

$$I = \frac{t^{19}}{360} \left[ 9\left(t + \frac{1}{t}\right) + \frac{1}{t} \right] + C$$

Substituting back $t + 1/t = 2\sqrt{1+x^2}$ and $1/t = \sqrt{1+x^2} - x$, we get:

$$I = \frac{(\sqrt{1+x^2}+x)^{19}}{360} \left[ 18\sqrt{1+x^2} + \sqrt{1+x^2} - x \right] + C$$

$$I = \frac{(\sqrt{1+x^2}+x)^{19}}{360} (19\sqrt{1+x^2} - x) + C$$

By comparing this to the target form, we find **$n = 19$** and **$m = 360$**. Thus, **$m+n = 379$**. You have conquered the monster!

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{( 1 + x ^{2} + x ) ^{10}}{( 1 + x ^{2} - x ) ^{9}} d x = \frac{1}{m} ((1 + x^{2} + x)^{n} (n 1 + x^{2} - x)) + C$ where $C$ is the constant of integration and $m, n \in N$ , then $m + n$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Integral Structure

The Strategic Substitution

The Conjugate Property

Isolating $1 + x^{2}$

Isolating $x$ in terms of $t$

Finding the Differential $d x$

Transforming the Integral

Algebraic Simplification

Performing the Integration

Factoring for the Target Form

Rearranging the Bracket

Back-Substitution to $x$

Final Form Comparison

Final Answer Calculation

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Conjugate Insight

The Strategic Substitution

The Transformation

The Final Alignment

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Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+x2​

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

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