If ∫ dx/(x²-2x+10)² = A (tan⁻¹((x-1)/3) + f(x)/(x²-2x+10)) + C, where C is a constant of integration

Q: Why can't we use partial fractions here?

Partial fractions are designed for rational functions where the denominator can be factored into linear or irreducible quadratic factors. Here, the denominator is an irreducible quadratic raised to the power of two, which makes partial fractions ineffective. We need a more robust method like trigonometric substitution.

Q: Why did we use $t = 3 \tan \theta$ instead of $t = 3 \sin \theta$?

Great question! We have $t^2 + 9$ in the denominator. If we used $\sin \theta$, we would get $9 \sin^2 \theta + 9 = 9 \cos^2 \theta$, which is fine, but $\tan \theta$ is the standard choice for $t^2 + a^2$ because it simplifies the differential $dt$ and the denominator term perfectly into a secant-based expression.

Q: How do I know when to stop and compare with the target form?

The target form is your map. Once you have an inverse tangent and a rational function, you know you've reached the destination. Just match the coefficients $A$ and the function $f(x)$ to the expression you derived. It's the final victory lap!

Question

If $\int \frac{dx}{(x^2-2x+10)^2} = A \left( 	an^{-1} \left( \frac{x-1}{3} ight) + \frac{f(x)}{x^2-2x+10} ight) + C$, where $C$ is a constant of integration, then :

Options:

(A) $A = 1/27$ and $f(x) = 9(x-1)$

(B) $A = 1/81$ and $f(x) = 3(x-1)$

(C) $A = 1/54$ and $f(x) = 9(x-1)^2$

(D) $A = 1/54$ and $f(x) = 3(x-1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to tackle an integral that might look like a daunting wall of algebra, but beneath the surface, it is a beautiful, structured dance of calculus.

We are looking at the integral:

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

When you see a quadratic in the denominator, your first instinct should always be to look for a perfect square. We can rewrite $x^2 - 2x + 10$ as $(x^2 - 2x + 1) + 9$, which is $(x-1)^2 + 3^2$.

By doing this, we have transformed a generic quadratic into a sum of squares. Now, our integral looks like:

$$I = \int \frac{dx}{((x-1)^2 + 3^2)^2}$$

### The Trigonometric Bridge

Now, let us simplify our life with a substitution. Let $t = x - 1$, which implies $dt = dx$.

Our integral becomes:

$$I = \int \frac{dt}{(t^2 + 3^2)^2}$$

Whenever you see $t^2 + a^2$, your mind should immediately jump to the tangent substitution. Let $t = 3 	an 	heta$, which implies $dt = 3 \sec^2 	heta \, d	heta$.

The denominator term $t^2 + 9$ becomes $9 	an^2 	heta + 9$, which simplifies to $9 \sec^2 	heta$. Substituting these into our integral, we get:

$$I = \int \frac{3 \sec^2 	heta \, d	heta}{(9 \sec^2 	heta)^2}$$

### The Dance of Identities

This is where the magic happens. The expression simplifies to:

$$I = \int \frac{3 \sec^2 	heta}{81 \sec^4 	heta} \, d	heta = \frac{1}{27} \int \frac{1}{\sec^2 	heta} \, d	heta$$

Since $\frac{1}{\sec^2 	heta} = \cos^2 	heta$, we are left with $\frac{1}{27} \int \cos^2 	heta \, d	heta$. We use the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ to proceed.

This gives us:

$$I = \frac{1}{54} \int (1 + \cos 2	heta) \, d	heta$$

The integration is now trivial:

$$I = \frac{1}{54} \left( 	heta + \frac{\sin 2	heta}{2} ight) + C$$

### Returning to Reality

We have the answer in terms of $	heta$, but we need it in terms of $x$. We know $	heta = 	an^{-1}(\frac{t}{3})$.

For $\sin 2	heta$, we use the identity $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$. Substituting $	an 	heta = \frac{t}{3}$, we get:

$$\sin 2	heta = \frac{2(t/3)}{1 + (t/3)^2} = \frac{6t}{t^2+9}$$

Thus, $\frac{\sin 2	heta}{2} = \frac{3t}{t^2+9}$. Finally, replacing $t$ with $x-1$, we arrive at the final result:

$$I = \frac{1}{54} \left( 	an^{-1}\left(\frac{x-1}{3}ight) + \frac{3(x-1)}{(x-1)^2+9} ight) + C$$

You have successfully conquered the integral!

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

Select Answer:

Visualized Solution

Analyze the Integral Structure

Completing the Square

Initial Substitution

Trigonometric Substitution

Simplify the Integral

Apply Trigonometric Identity

Perform Integration

Convert Back to Tangent

Substitute Back for $t$

Final Back-Substitution

Conclusion and Final Answer

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Trigonometric Bridge

The Dance of Identities

Returning to Reality

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫(x2−2x+10)2dx​=A(tan−1(3x−1​)+x2−2x+10f(x)​)+C, where C is a constant of integration, then :