Q.Integrate the following function: $e^{2x+3}$

We have an integral of the form $\int e^{ax} (\sin bx \pm \cos bx)\,dx$. The standard approach is to assume the antiderivative is itself a linear combination of $e^{ax}\sin bx$ and $e^{ax}\cos bx$, then differentiate and match coefficients. This avoids the tedium of two rounds of integration by parts.

1. Set up the guess Since the derivative of $e^{4x}\sin 3x$ will produce both $e^{4x}\sin 3x$ and $e^{4x}\cos 3x$ (and similarly for $e^{4x}\cos 3x$), we assume

where $A$ and $B$ are constants to be found, and $C$ is the constant of integration.

2. Differentiate the guess Differentiate $e^{4x}(A\sin 3x + B\cos 3x)$ using the product rule:

Group the $\sin 3x$ and $\cos 3x$ terms:

3. Match with the integrand The integrand is $e^{4x}(\sin 3x - \cos 3x)$. So we require:

4. Solve the system From the first equation: $4A = 1 + 3B \Rightarrow A = \frac{1+3B}{4}$. Substitute into the second:

Multiply through by 4: $16B + 3 + 9B = -4 \implies 25B = -7 \implies B = -\frac{7}{25}$.

Then $A = \frac{1 + 3(-7/25)}{4} = \frac{1 - 21/25}{4} = \frac{4/25}{4} = \frac{1}{25}$.

5. Write the antiderivative So

But the given options have $7\sin 3x - \cos 3x$ or similar. Wait — check the sign: our result is $\frac{e^{4x}}{25}(\sin 3x - 7\cos 3x)$. That is not directly among the options. Let’s re-check the matching step.

Differentiate $\frac{e^{4x}}{25}(\sin 3x - 7\cos 3x)$:

It works! So our result is correct. But option (A) is $7\sin 3x - \cos 3x$, not $\sin 3x - 7\cos 3x$. Wait — are they the same? No, they are different unless we misread the options. Let’s check option (A) carefully:

(A) $\dfrac{e^{4x}}{25} (7 \sin 3x - \cos 3x) + c$

Our result: $\dfrac{e^{4x}}{25} (\sin 3x - 7 \cos 3x) + c$ …

Concept & Intuition

The integrand looks messy, but the square root contains two terms that share a factor of $1/x^2$. Factoring that out reveals a derivative of $(\log x)^2$ hiding inside. The key is to notice that

which appears almost exactly in the integrand. This suggests a substitution $u = 1+(\log x)^2$, turning the integral into a simple power rule.

Step-by-step solution

1. Simplify the square root

(We take the positive root since $x>0$ for the log.)

2. Rewrite the integral

3. Spot the derivative Let $t = 1 + (\log x)^2$. Then

The integral becomes

4. Integrate

Substituting back $t = 1+(\log x)^2$:

5. Match the given form The problem states the integral equals $\frac{f(x)}{3} \sqrt{1+(\log x)^2} + c$. Our result is $\frac13 \bigl(1+(\log x)^2\bigr)^{3/2}$. …

The problem gives you an integral already expressed in a partially solved form, with an unknown function $f(x)$ inside. Your job is to figure out what $f(x)$ must be, then evaluate it at $x = 0$. The structure of the answer — terms like $\frac{1}{3}[f(x)]^{3/2}$ and $\sqrt{f(x)}$ — strongly suggests that $f(x)$ is the expression inside the square root of the original integrand. Let’s see why.

1. Identify the natural candidate for $f(x)$. The integrand is $(e^{2x} + 2e^x) \sqrt{e^{2x} - 4e^x + 5}$. The expression under the square root, $e^{2x} - 4e^x + 5$, is the most likely candidate for $f(x)$, because it appears inside $\sqrt{\cdot}$ in the integrand and also appears in the answer as $\sqrt{f(x)}$ and $[f(x)]^{3/2}$. So we guess:

2. Verify by performing the integration. Let $t = e^x - 2$. Then $dt = e^x \, dx$, and also $e^x = t + 2$. Compute $e^{2x} = (t+2)^2 = t^2 + 4t + 4$. Then:

So $\sqrt{e^{2x} - 4e^x + 5} = \sqrt{t^2 + 1}$.

Now the other factor: $e^{2x} + 2e^x = (t+2)^2 + 2(t+2) = t^2 + 4t + 4 + 2t + 4 = t^2 + 6t + 8$.

The integrand becomes $(t^2 + 6t + 8) \sqrt{t^2 + 1}$, and $dx = \frac{dt}{e^x} = \frac{dt}{t+2}$.

So the integral is:

Factor $t^2 + 6t + 8 = (t+2)(t+4)$. Cancel $t+2$:

3. Split and integrate.

For the first integral, substitute $u = t^2 + 1$, $du = 2t \, dt$, so $t \, dt = \frac{du}{2}$:

For the second integral, recall the standard formula: …

The key here is that the equation is already in the simplest possible form: $\frac{dy}{dx}$ is given as a function of $x$ alone. There is no $y$ on the right-hand side, so this is not a separable or linear equation in the usual sense — it’s just a direct integration problem. The derivative of $y$ with respect to $x$ is known, so $y$ itself is the antiderivative (indefinite integral) of $e^{-2x}$, plus a constant. The initial condition $y(\log 2) = \frac{1}{16}$ then pins down that constant.

Let’s go through it step by step.

1. Integrate both sides We have

Integrating with respect to $x$:

The integral of $e^{-2x}$ is $-\frac{1}{2} e^{-2x}$, because the derivative of $e^{-2x}$ is $-2e^{-2x}$, so we divide by $-2$ to reverse it. Thus

where $C$ is the constant of integration.

2. Apply the initial condition We are told that when $x = \log 2$, $y = \frac{1}{16}$. Substitute:

Simplify $e^{-2 \log 2}$. Since $e^{\log 2} = 2$, we have $e^{-2 \log 2} = (e^{\log 2})^{-2} = 2^{-2} = \frac{1}{4}$. So

Therefore

3. Write the particular solution Substitute $C = \frac{3}{16}$ back: $y = -\frac{1}{2} e^{-2x} + \frac{3}{16}.$ …