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

Step 1 — Why this form is the right tool.

An integrand shaped like $e^{x}\big[f(x) + f'(x)\big]$ integrates in one line to $e^{x}f(x)$, because $\frac{d}{dx}\big(e^{x}f(x)\big) = e^{x}f(x) + e^{x}f'(x)$. Our job is to massage $\dfrac{x e^{x}}{(2+x)^{3}}$ into that shape.

Step 2 — Split the numerator.

Write $x = (2+x) - 2$:

Step 3 — Recognise $f$ and $f'$.

Take

So the bracket is precisely $f(x) + f'(x)$, and

Step 4 — Apply the limits $0$ to $1$. …

The key insight: when you see a product of a logarithm and a rational function, integration by parts is almost always the way. The logarithm’s derivative is simple ($1/x$), and the rational part often integrates nicely. Here, $(1+x)^{-2}$ is the derivative of $-1/(1+x)$, so letting $u = \log x$ and $dv = dx/(1+x)^2$ is natural.

Let’s work through it step by step.

1. Set up integration by parts. Let

Then

(Check: derivative of $-1/(1+x)$ is $1/(1+x)^2$.)

2. Apply the formula $\int u\,dv = uv - \int v\,du$:

The minus signs give:

3. Simplify the remaining integral using partial fractions. Write

Multiply through by $x(1+x)$:

For $x = 0$: $1 = A$.

For $x = -1$: $1 = -B \Rightarrow B = -1$.

So

4. Integrate term by term:

Combine the logs:

5. Put it all together: $$$$

The key insight is that $\log x^2$ is not $(\log x)^2$; it's $2\log x$. This simplifies the integral into a standard form that integration by parts handles cleanly. Many students mistakenly try to integrate $\log x^2$ as if it were a square, but the logarithm's power rule saves the day.

1. Rewrite the integrand Using the logarithm property $\log a^b = b\log a$, we have:

So the integral becomes:

2. Integrate $\int \log x \, dx$ by parts Recall the integration by parts formula: $\int u\,dv = uv - \int v\,du$. Choose:

Then:

So:

3. Multiply by 2 and express in original form $2\int \log x \, dx = 2(x\log x - x) + C = 2x\log x - 2x + C$ …

We start by noticing that the integrand is $\log x (\log x + 2) = (\log x)^2 + 2\log x$. This looks like something that might be the derivative of a product involving $\log x$. A classic trick: the derivative of $x (\log x)^n$ gives terms like $n (\log x)^{n-1} + (\log x)^n$. Here, if we try $n=2$, we get exactly the pattern we need.

Let’s verify step by step.

1. Simplify the integrand

2. Guess a candidate antiderivative Consider $F(x) = x (\log x)^2$. Differentiate using the product rule:

This matches exactly the integrand.

3. Conclude the indefinite integral Since $F'(x)$ equals the integrand, we have $\int \left[ (\log x)^2 + 2\log x \right] dx = x (\log x)^2 + C.$ …

Recall the standard result

We must split $\dfrac{x^2+1}{(x+1)^2}$ as $f(x)+f'(x)$. Try

Then

Adding: …

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: the standard form integral of e^x [ f(x) + f'(x) ] dx = e^x f(x) + c.

Rewrite the bracket using 1 + tan^2 x = sec^2 x:

1 + tan x + tan^2 x = tan x + (1 + tan^2 x) = tan x + sec^2 x.

So the integral is

integral e^x [ tan x + sec^2 x ] dx.

Here f(x) = tan x and f'(x) = sec^2 x, exactly the required pattern. …

Step 1 — The concept

The standard result comes straight from the product rule:

So whenever an integrand is $e^x$ times (some function + its own derivative), the answer is simply $e^x f(x) + c$. Our job is to force $\dfrac{x}{(1+x)^2}$ into that shape.

Step 2 — Split the rational part

Write the numerator as $x = (1+x) - 1$:

Step 3 — Identify $f$ and $f'$

Take

Then exactly

Step 4 — Integrate …

The key insight here is that the integrand $e^x \cdot \frac{1+\sin x}{1+\cos x}$ is a product of $e^x$ with a trigonometric expression. Integrals of the form $\int e^x [f(x) + f'(x)]\,dx$ have a neat closed form: $e^x f(x) + c$. So if we can rewrite $\frac{1+\sin x}{1+\cos x}$ as $f(x) + f'(x)$ for some function $f(x)$, the integral becomes trivial.

Let’s see if that works.

1. Simplify the trigonometric fraction. Use the half-angle identities: $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$, $\cos x = 2\cos^2\frac{x}{2} - 1 = 1 - 2\sin^2\frac{x}{2}$, and $1 + \cos x = 2\cos^2\frac{x}{2}$. Then

2. Split the numerator.

3. Recognise the $f(x) + f'(x)$ pattern. Let $f(x) = \tan\frac{x}{2}$. Then

So indeed

4. Apply the standard result. The integral becomes …

Use $\int e^x\big(f(x)+f'(x)\big)dx=e^x f(x)+c$. Take $f(x)=\dfrac{1}{1+x^2}$, so $f'(x)=\dfrac{-2x}{(1+x^2)^2}$. Then

$f+f'=\frac{(1+x^2)-2x}{(1+x^2)^2}=\frac{(1-x)^2}{(1+x^2)^2}=\left(\frac{1-x}{1+x^2}\right)^2,$ …

Step 1 — First integration by parts.

With $u=x^{3}$, $dv=\sin 3x\,dx \Rightarrow v=-\dfrac{\cos 3x}{3}$:

Step 2 — Second integration by parts.

With $u=x^{2}$, $dv=\cos 3x\,dx \Rightarrow v=\dfrac{\sin 3x}{3}$:

Step 3 — Third integration by parts.

With $u=x$, $dv=\sin 3x\,dx \Rightarrow v=-\dfrac{\cos 3x}{3}$:

Step 4 — Back-substitute into Step 2.

Step 5 — Back-substitute into Step 1.

Step 6 — Verify by differentiation (spot-check). …