Q.Prove that $\int_{0}^{1}x\,e^x\,dx=1$

Standard results (by integration by parts):

Subtract: …

We start with the integral

Concept & Intuition

The argument $\sqrt{\frac{1+x^2}{1-x^2}}$ looks like a tangent of some angle. Recall that $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$. If we set $x^2 = \cos 2\theta$, then $1+x^2 = 1+\cos2\theta = 2\cos^2\theta$ and $1-x^2 = 1-\cos2\theta = 2\sin^2\theta$. Their ratio becomes $\cot^2\theta$, so the square root is $\cot\theta$. Then $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$. This transforms the messy inverse tangent into a simple linear function of $\theta$.

Step-by-step solution

1. Substitute $x^2 = \cos 2\theta$ Let $x^2 = \cos 2\theta$, so that $0 \le 2\theta \le \pi$ (to keep things principal). Then

Hence

Therefore

2. Rewrite the integral in terms of $\theta$ From $x^2 = \cos 2\theta$, differentiate: $2x\,dx = -2\sin 2\theta\,d\theta$ → $x\,dx = -\sin 2\theta\,d\theta$. Also $\sin 2\theta = 2\sin\theta\cos\theta$. So

3. Integrate by parts Let $u = \frac{\pi}{2} - \theta$ and $dv = \sin 2\theta\,d\theta$. Then $du = -d\theta$ and $v = -\frac{1}{2}\cos 2\theta$. Integration by parts:

So

Simplify:

The integral of $\cos 2\theta$ is $\frac{1}{2}\sin 2\theta$. Thus

4. Return to $x$ Recall $I = -$ of the above, so $I = \frac{1}{2}\left( \frac{\pi}{2} - \theta \right)\cos 2\theta + \frac{1}{4}\sin 2\theta + C.$ …

We start with

Concept & Intuition

The expression inside the square root, $\frac{1+x^2}{1-x^2}$, looks like something trigonometric. If we set $x^2 = \cos 2\theta$ (or equivalently $x = \sqrt{\cos 2\theta}$), then

Thus $\sqrt{\frac{1+x^2}{1-x^2}} = \cot\theta$, and $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$. This turns the messy inverse tangent into a simple linear function of $\theta$. Then we change variables from $x$ to $\theta$ and integrate.

Step‑by‑step solution

1. Substitute $x^2 = \cos 2\theta$. Let $0 \le 2\theta \le \pi$ so that $x^2 \in [-1,1]$ (the domain of the square root). Then

Hence

2. Simplify the inverse tangent. For $\theta \in (0,\pi/2)$, $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$. Therefore

3. Change the differential. From $x^2 = \cos 2\theta$, differentiate:

So the integral becomes

4. Integrate with respect to $\theta$. Let $u = \frac{\pi}{2} - \theta$, then $du = -d\theta$ and $\sin 2\theta = \sin(\pi - 2u) = \sin 2u$. Also $\theta = \frac{\pi}{2} - u$. Alternatively, integrate directly by parts:

Set $A = \frac{\pi}{2} - \theta$, $dB = \sin 2\theta\,d\theta$. Then $dA = -d\theta$, $B = -\frac{1}{2}\cos 2\theta$.

Integration by parts gives

Simplify carefully:

The minus sign outside flips signs:

Now $\int \cos 2\theta\,d\theta = \frac{1}{2}\sin 2\theta$. So

We are told that

and we need $27\bigl(f(x) + x g(x)\bigr)$. The key idea: differentiate both sides to eliminate the integral, then compare coefficients of $\cos 3x$ and $\sin 3x$ to solve for $f(x)$ and $g(x)$.

1. Differentiate both sides Since the derivative of the integral is the integrand, we have

Use the product rule on each term:

Adding them:

2. Match coefficients The left side has no $\cos 3x$ term, so its coefficient must be zero:

The coefficient of $\sin 3x$ on the left is $x^3$, so:

3. Solve the system From (1): $f'(x) = -3g(x)$. Differentiate (2):

Substitute $f'(x) = -3g(x)$:

This is a second-order linear ODE. The homogeneous solution is $g_h(x) = A\cos 3x + B\sin 3x$, but since $f$ and $g$ are likely polynomials (the integrand is a polynomial times sine), we try a particular solution of the form $g_p(x) = ax^2 + b$.

Then $g_p'' = 2a$, so:

Matching: $9a = 3 \Rightarrow a = \frac{1}{3}$, and $2a + 9b = 0 \Rightarrow \frac{2}{3} + 9b = 0 \Rightarrow b = -\frac{2}{27}$.

So $g(x) = \frac{1}{3}x^2 - \frac{2}{27}$ (ignoring homogeneous part, as it would introduce extra trig terms not present in the given form).

4. Find $f(x)$ From (1): $f'(x) = -3g(x) = -3\left(\frac{1}{3}x^2 - \frac{2}{27}\right) = -x^2 + \frac{2}{9}$. Integrate: $f(x) = -\frac{x^3}{3} + \frac{2}{9}x + C$. The constant $C$ would produce a term $C\cos 3x$ in the antiderivative, but the integral of $x^3\sin 3x$ has no pure cosine term (check by differentiating: it would give a sine term), so $C = 0$. Thus

5. Compute the required expression We need $27\bigl(f(x) + x g(x)\bigr)$. …

Concept & Intuition

When integrating powers of $\log x$, the key trick is to treat $\log x$ as the function to differentiate and $1$ (or $dx$) as the function to integrate. Why? Because the derivative of $\log x$ is $1/x$, which cancels the $x$ that appears when we integrate $1$. This lets us reduce the exponent of the logarithm step by step. It’s like peeling an onion: each integration by parts lowers the power by one, until we’re left with a simple integral.

Step-by-step solution

1. Set up integration by parts Let $u = (\log x)^3$ and $dv = dx$. Then $du = 3(\log x)^2 \cdot \frac{1}{x}\, dx$ and $v = x$. The formula $\int u\,dv = uv - \int v\,du$ gives:

2. Reduce the power again Now we need $\int (\log x)^2\,dx$. Use the same trick: let $u = (\log x)^2$, $dv = dx$. Then $du = 2(\log x)\cdot\frac{1}{x}\,dx$, $v = x$.

3. Handle $\int \log x\,dx$ This is a classic: let $u = \log x$, $dv = dx$. Then $du = \frac{1}{x}dx$, $v = x$.

4. Substitute back From step 2:

From step 1:

$\int (\log x)^3\,dx = x(\log x)^3 - 3\left[ x(\log x)^2 - 2x\log x + 2x \right] + c.$ …

Step 1 — Integration by parts

Take $u = \ln x$, $dv = dx$, so $du = \dfrac{1}{x}dx$, $v = x$:

Step 2 — Complete the integral

Step 3 — Rewrite in the option's form

Factor x and use $1 = \ln e$:

Why the other options are wrong: (A) x·ln(ex) = x(ln x + 1) has the wrong sign; (C) x·ln(e/x) reverses the ratio; (D) drops the leading x factor. Only (B) matches x ln x − x. …

Concept & Intuition

When integrating a product of a polynomial and a logarithm, integration by parts is the natural tool. Here $x^4$ is easy to integrate and $(\log x)^3$ becomes simpler when differentiated. Repeating the process three times reduces the power of $\log x$ to zero, leaving a pure polynomial integral. The final expression matches the given form, so we can read off $A, B, C, D$ by comparing coefficients.

Step-by-step solution

1. Set up integration by parts Let $u = (\log x)^3$ and $dv = x^4\,dx$. Then

Integration by parts gives

2. Second integration by parts Now handle $\int x^4 (\log x)^2\,dx$. Let $u = (\log x)^2$, $dv = x^4\,dx$. Then

So

Substituting back:

3. Third integration by parts For $\int x^4 \log x\,dx$, let $u = \log x$, $dv = x^4\,dx$. Then

Hence

Substitute this into the previous expression:

4. Identify coefficients The given form is

Matching term by term:

$A = \frac{1}{5},\quad B = -\frac{3}{25},\quad C = \frac{6}{125},\quad D = -\frac{6}{625}.$ …

The key idea: $\left|\log_{10} x\right|$ changes sign at $x=1$ because $\log_{10}1=0$. For $x<1$, the logarithm is negative, so the absolute value flips its sign; for $x>1$, it stays positive. So we split the integral at $x=1$ and integrate each piece separately, remembering that $\log_{10} x = \frac{\ln x}{\ln 10}$.

1. Find where the integrand changes behavior

   $\log_{10} x = 0 \implies x=1$.

   On $[\frac12,1]$, $\log_{10} x \le 0$, so $|\log_{10} x| = -\log_{10} x$.

   On $[1,2]$, $\log_{10} x \ge 0$, so $|\log_{10} x| = \log_{10} x$.

2. Write the integral as a sum

3. Convert to natural logs for easy integration $\log_{10} x = \frac{\ln x}{\ln 10}$, so

4. Integrate $\ln x$ Recall $\int \ln x\,dx = x\ln x - x + C$.
   • For the first integral:

• For the second integral:

5. Combine the pieces $I = \frac{1}{\ln 10}\left( -\left( \frac12 \ln 2 - \frac12 \right) + (2\ln 2 - 1) \right) = \frac{1}{\ln 10}\left( -\frac12 \ln 2 + \frac12 + 2\ln 2 - 1 \right) = \frac{1}{\ln 10}\left( \frac32 \ln 2 - \frac12 \right).$ …

Concept and Intuition

When you see a product of a polynomial and a power of $\log x$, integration by parts is the natural tool: each application reduces the power of $\log x$ by one, until only a pure polynomial integral remains. The given form is that result, factored so its coefficients are integers — we just need to integrate carefully and match coefficients.