Q.Find the following integrals:

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. The other options do not differentiate to the given integrand (quick check: differentiate $x \cdot 3^x$ gives $3^x + x \cdot 3^x \log 3$, not matching; $\frac{3^x}{x^2}$ differentiates to something with $x^3$ in denominator; $x^2 3^x$ gives a different combination).

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

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

The $(1-t)^2$ cancels, and $\frac{3}{9} = \frac{1}{3}$, so we get

5. Integrate using the power rule

Thus

6. Substitute back $t = \frac{x-1}{x+2}$

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.

(Check: differentiating $\frac{7}{4}t^{2/7}$ gives $\frac{1}{2}t^{-5/7}\cdot\frac{2}{(x+1)^2} = \frac{(x+1)^{5/7}}{(x-1)^{5/7}(x+1)^2}$, which is the original integrand.)

ANSWER: C

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 $t$

Step 4 — Return to $x$ and check

Sanity check with $m=2$: the formula gives $-\dfrac{\sqrt{x^{2}+1}}{x}+c$, and differentiating that indeed returns $\dfrac{1}{x^{2}\sqrt{x^{2}+1}}$ — the original integrand for $m=2$.

ANSWER: B

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:

The $g(x)$ cancels neatly:

5. Evaluate the integral Let $u = \log f(x)$. Then $du = \frac{f'(x)}{f(x)} dx$. The integral becomes:

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 $H(x) = \frac{g(x)}{f(x)}$: