Animated Solution for Mathematics - Indefinite Integration: Integrate $\int \frac{x^3+3x+2}{(x^2+1{)}^2(x+1)}dx$

Unraveling the Rational Function

Welcome, future engineer. Today, we are not just solving an integral; we are performing a surgical operation on a complex rational function.

When you look at the expression

I want you to see past the intimidating powers. I want you to see a puzzle waiting to be dismantled.

The degree of our numerator is 3, and the degree of our denominator is 5. Because the numerator is of a lower degree, we are in the perfect position to use the Method of Partial Fractions. This is our primary tool for breaking down complex, high-degree rational functions into manageable, bite-sized pieces.

The Blueprint

We begin by setting up our decomposition. We have a linear factor $(x+1)$ and a repeated quadratic factor $(x^2+1{)}^2$.

The rules of partial fractions dictate that for every linear factor, we assign a constant, and for every irreducible quadratic factor, we assign a linear numerator. Because our quadratic factor is repeated, we must account for both the first power and the second power.

Thus, we write:

This is our blueprint. Our mission now is to find the constants $A,B,C,D,$ and $E$.

The Algebraic Hunt

To find these constants, we multiply the entire equation by the denominator $(x^2+1{)}^2(x+1)$. This gives us the identity:

This identity is our strongest weapon; it holds true for any value of $x$. By choosing $x=-1$, we instantly eliminate the terms involving $(x+1)$, allowing us to isolate $A$. We find that $A=-\frac{1}{2}$.

With $A$ in hand, we compare coefficients of $x^4,x^3,x^2,$ and the constant term. By equating coefficients, we discover $B=\frac{1}{2}$, $C=\frac{1}{2}$, $D=0$, and $E=2$. The math is elegant; the coefficients fall into place like tumblers in a lock.

The Integration Strategy

With our constants found, the integral splits into three distinct parts:

The first two integrals are standard. The first, $\int \frac{-\frac{1}{2}}{x+1}dx$, yields $-\frac{1}{2}\log |x+1|$.

The second, $\int \frac{\frac{1}{2}x+\frac{1}{2}}{x^2+1}dx$, splits into a logarithmic term and an inverse tangent term: $\frac{1}{4}\log (x^2+1)+\frac{1}{2}{\tan }^{-1}x$. These are the bread and butter of calculus.

The Trigonometric Pivot

We are left with $I_3=2\int \frac{1}{(x^2+1{)}^2}dx$. Whenever you see $(x^2+a^2)$ in the denominator, especially raised to a power, think of trigonometric substitution.

Let $x=\tan \theta$. Then $dx={\sec }^2\theta d\theta$. The integral transforms beautifully:

Using the identity $2{\cos }^2\theta =1+\cos 2\theta$, we integrate to get $\theta +\frac{1}{2}\sin 2\theta$. Back-substituting $\theta ={\tan }^{-1}x$ and using the triangle identity for $\sin 2\theta =\frac{2x}{1+x^2}$, we arrive at ${\tan }^{-1}x+\frac{x}{1+x^2}$.

We combine all our pieces:

Combining the ${\tan }^{-1}x$ terms gives us the final, elegant result:

You have navigated the decomposition, the algebraic comparison, and the trigonometric substitution. This is the essence of JEE Advanced mathematics—a journey of logical steps leading to a beautiful, unified solution.