Q.Integrate the following function: $\int (1 - x) \sqrt{x} dx$

The concept first

Integrals of the type $\displaystyle\int\frac{dx}{x^{a}\left(x^{m}+1\right)^{b}}$ rarely yield to the naive substitution $u=x^{m}+1$. The standard trick is to take the highest power out of the bracket: $x^{m}+1=x^{m}\left(1+x^{-m}\right)$. After that, every $x$ appears through the single combination $1+x^{-m}$, and one substitution finishes the job.

Step 1 — Rewrite the radical

For $x>0$,

Hence

Step 2 — Substitute

Let

Differentiating both sides: $m\,t^{m-1}\,dt=-m\,x^{-m-1}\,dx$, i.e.

Step 3 — Integrate in $t$ …

The key insight is that the integrand has the form $(x-1)^{-3/4}(x+2)^{-5/4}$. When the exponents are fractions, a common trick is to rewrite the product as a single factor raised to a power times a rational function of a ratio. Here, notice that the sum of the exponents is $-3/4 - 5/4 = -2$, which suggests factoring out $(x+2)^{-2}$ and then expressing the rest in terms of $\frac{x-1}{x+2}$.

1. Rewrite the integrand Factor $(x+2)^{-5/4}$ as $(x+2)^{-2} \cdot (x+2)^{3/4}$:

But it’s more convenient to work with the reciprocal ratio. Let

Then

So the integrand becomes

2. Find $dx$ in terms of $t$ From $t = \frac{x-1}{x+2}$, solve for $x$:

So

Differentiate with respect to $t$:

Hence

3. Express $(x+2)^2$ in terms of $t$ Since $x = \frac{1+2t}{1-t}$, we have

Therefore,

4. Substitute everything into the integral The integral becomes $$$$

We are asked to integrate

The presence of $3^x$ and a rational function suggests the derivative of something like $\frac{3^x}{x}$ might appear, because the derivative of $3^x$ is $3^x \log 3$, and the quotient rule naturally produces a term like $x \log 3 - 1$ in the numerator.

Let’s check:

1. Recall the derivative of $a^x$

Here $a=3$, so $\frac{d}{dx} 3^x = 3^x \log 3$.

2. Consider the function $f(x) = \frac{3^x}{x}$ Differentiate using the quotient rule:

This is exactly the integrand we have.

3. Therefore, the integral is immediate $$$$

Concept. For $\displaystyle\int \frac{dx}{(x-a)^{p}(x+b)^{q}}$ with $p + q = 2$, the substitution $t = \dfrac{x-a}{x+b}$ works, because $\dfrac{dt}{dx} = \dfrac{(a+b)}{(x+b)^2}$ supplies exactly the leftover factor.

Step 1 — rewrite the integrand. Here $\frac{5}{7} + \frac{9}{7} = 2$:

Step 2 — substitute $t = \dfrac{x-1}{x+1}$, so $dt = \dfrac{(x+1)-(x-1)}{(x+1)^2}dx = \dfrac{2}{(x+1)^2}dx$:

Step 3 — back-substitute. …

We are asked to find

and choose among four options. The key is to recognize that the integrand looks like the derivative of a quotient of the form $\frac{3^x}{x}$.

Why this approach works:

When we see $3^x$ and a denominator $x^2$, a natural guess is $\frac{3^x}{x}$ because its derivative will involve $3^x \log 3$ (from differentiating the exponential) and a $1/x^2$ term (from differentiating $1/x$). The given numerator $x \log 3 - 1$ is exactly what appears when we apply the quotient rule to $\frac{3^x}{x}$.

Let’s verify step by step.

1. Differentiate $\frac{3^x}{x}$ Write $3^x = e^{x \log 3}$. Then

by the quotient rule (or product rule: $3^x \cdot x^{-1}$).

Factor $3^x$ from the numerator:

2. Compare with the integrand The derivative we just computed is exactly the integrand. Therefore,

3. Match with the options Option (D) is $\frac{3^x}{x} + c$, which matches our result. …

Concept and Intuition

The integrand is $\frac{x}{(2-x)^4}$. As $x$ approaches $2$ from the left, the denominator $(2-x)^4$ goes to $0$, and the whole fraction blows up to $+\infty$. This is an improper integral because the function is unbounded at the upper limit. The correct approach is to replace the upper bound with a variable $t$, integrate, and then take the limit $t \to 2^-$. If that limit is infinite, the integral diverges.

A common pitfall is to treat it as a normal definite integral and apply the power rule without considering the singularity — that leads to a finite (but wrong) answer. The problem is designed to catch that mistake.

Step-by-step solution

1. Recognize the improper nature The integrand $\frac{x}{(2-x)^4}$ has a vertical asymptote at $x=2$ (the denominator is zero). Since the upper limit of integration is exactly $2$, we must write:

2. Substitute to simplify Let $u = 2 - x$. Then $x = 2 - u$, and $dx = -du$. When $x = 0$, $u = 2$; when $x = t$, $u = 2 - t$. The integral becomes:

(The minus sign flips the limits.)

3. Split the integrand

So the integral is:

4. Integrate term by term

So:

5. Evaluate from $u = 2-t$ to $u = 2$

At $u=2$:

At $u = 2-t$:

So the definite integral from $0$ to $t$ is:

We start with the function $y = f(x)^{g(x)}$. The derivative is given in a special form:

Our task is to find $\int \frac{G(x) H(x) f'(x)}{g(x)} dx$. The key is to determine what $H(x)$ and $G(x)$ actually are by computing the derivative of $y$ directly using logarithmic differentiation.

Why logarithmic differentiation?

When a variable appears in both the base and the exponent, the standard differentiation rules (power rule, exponential rule) don’t apply directly. Taking logs converts the exponent into a product, making differentiation straightforward.

1. Take the natural logarithm of both sides

2. Differentiate implicitly Differentiate both sides with respect to $x$:

Multiply through by $y$:

3. Match with the given form The problem states:

Comparing term-by-term, we identify:

4. Substitute into the required integral We need: $\int \frac{G(x) H(x) f'(x)}{g(x)} \, dx = \int \frac{(\log f(x)) \cdot \frac{g(x)}{f(x)} \cdot f'(x)}{g(x)} \, dx.$ …

When you see a function raised to another function — $f(x)^{g(x)}$ — the standard tool is logarithmic differentiation. Take logs, differentiate implicitly, and you get an expression of the form $\frac{dy}{dx} = y \left[ g'(x) \log f(x) + \frac{g(x) f'(x)}{f(x)} \right]$. The problem gives a similar form but with $H(x)$ and $G(x)$ as placeholders. Our job is to match them, then evaluate the integral they ask for.

Let’s work through it step by step.

1. Start with logarithmic differentiation. Given $y = f(x)^{g(x)}$, take natural logs:

Differentiate both sides with respect to $x$:

Multiply through by $y$:

2. Match with the given form. The problem states:

Comparing term-by-term with our derived expression:

• The term with $f'(x)$ is $\frac{g(x)}{f(x)} f'(x)$, so $H(x) = \frac{g(x)}{f(x)}$.
• The term with $g'(x)$ is $\log f(x) \cdot g'(x)$, so $G(x) = \log f(x)$.

3. Set up the required integral. We need:

Substitute $G(x) = \log f(x)$ and $H(x) = \frac{g(x)}{f(x)}$:

The $g(x)$ cancels:

4. Simplify the integral. So the integral becomes:

Notice that $f'(x) \, dx = d(f(x))$. Let $u = f(x)$, then $du = f'(x) dx$, and the integral is: …