Q.Show that $\int \frac{(a^x+b^x)^2}{a^x b^x}\,dx = \frac{\left(\frac{a}{b}\right)^x-\left(\frac{b}{a}\right)^x}{\log a-\log b}+2x+C$.

Polynomial Integration — From Intuition to Precision

Think of integration as the reverse of differentiation. If differentiation tells you the slope of a curve at every point, integration tells you the area under that curve between two points. For polynomials, this reverse process is beautifully simple.

The Intuition: Undoing the Power Rule

You already know the power rule for differentiation: if $f(x) = x^n$, then $f'(x) = n x^{n-1}$. Integration asks: given that the derivative is $x^n$, what was the original function?

Suppose you want a function whose derivative is $x^2$. You need something that, when differentiated, gives $x^2$. Try $x^3$: its derivative is $3x^2$, which is three times too big. So try $\frac{1}{3}x^3$ — its derivative is exactly $x^2$. That's the core idea: increase the exponent by 1, then divide by the new exponent.

The $+C$ is crucial. Why? Because the derivative of any constant is zero. If $F(x) = \frac{1}{3}x^3 + 5$, its derivative is still $x^2$. So when we integrate, we must add an arbitrary constant $C$ to account for all possible original functions.

The Precise Statement

For a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, its indefinite integral (antiderivative) is:

You integrate term by term, applying the reverse power rule to each term separately. The constant term $a_0$ integrates to $a_0 x$, since $\int a_0 \, dx = a_0 x$ (because the derivative of $a_0 x$ is $a_0$).

A Worked Example

Integrate $f(x) = 4x^3 - 2x + 7$.

Apply the rule term by term:

• $4x^3$: increase exponent to 4, divide by 4 → $\frac{4}{4}x^4 = x^4$
• $-2x$: this is $-2x^1$, increase exponent to 2, divide by 2 → $\frac{-2}{2}x^2 = -x^2$
• $7$: this is $7x^0$, increase exponent to 1, divide by 1 → $7x$

So:

You can check by differentiating: the derivative of $x^4 - x^2 + 7x + C$ is $4x^3 - 2x + 7$, which is exactly your original function.

Definite Integration: Area Under the Curve

When you want the actual area between $x = a$ and $x = b$, you use the definite integral: …