Maharashtra Board Class 12 Maths Part 1 Chapter 5 Integration 5.5 Solutions

Evaluate The Following.

Question 1. \( \int x \log x \, dx \)
Answer: Solution:
Let \( I = \int x \log x \, dx \)
Using Integration by Parts: \( \int u \, dv = uv - \int v \, du \)
Let \( u = \log x \implies du = \frac{1}{x} \, dx \)
Let \( dv = x \, dx \implies v = \int x \, dx = \frac{x^2}{2} \)
\( I = (\log x) \left( \frac{x^2}{2} \right) - \int \left( \frac{x^2}{2} \right) \left( \frac{1}{x} \right) \, dx \)
\( = \frac{x^2}{2} \log x - \int \frac{x}{2} \, dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \left( \frac{x^2}{2} \right) + c \)
\( = \frac{x^2}{2} \log x - \frac{x^2}{4} + c \)
In simple words: This question asks us to integrate the product of x and log x. We solve it using integration by parts, treating log x as the first function and x as the second function, following the LIATE rule.

🎯 Exam Tip: Remember the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) for choosing the first function in integration by parts to ensure efficient calculation.

Question 2. \( \int x^2 e^{4x} \, dx \)
Answer: Solution:
Let \( I = \int x^2 e^{4x} \, dx \)
Using Integration by Parts: \( \int u \, dv = uv - \int v \, du \)
Let \( u = x^2 \implies du = 2x \, dx \)
Let \( dv = e^{4x} \, dx \implies v = \int e^{4x} \, dx = \frac{e^{4x}}{4} \)
\( I = x^2 \left( \frac{e^{4x}}{4} \right) - \int \left( \frac{e^{4x}}{4} \right) (2x) \, dx \)
\( = \frac{x^2 e^{4x}}{4} - \frac{1}{2} \int x e^{4x} \, dx \)
Now, we need to integrate \( \int x e^{4x} \, dx \) again using integration by parts.
Let \( u = x \implies du = 1 \, dx \)
Let \( dv = e^{4x} \, dx \implies v = \frac{e^{4x}}{4} \)
So, \( \int x e^{4x} \, dx = x \left( \frac{e^{4x}}{4} \right) - \int \left( \frac{e^{4x}}{4} \right) (1) \, dx \)
\( = \frac{x e^{4x}}{4} - \frac{1}{4} \int e^{4x} \, dx \)
\( = \frac{x e^{4x}}{4} - \frac{1}{4} \left( \frac{e^{4x}}{4} \right) + c_1 \)
\( = \frac{x e^{4x}}{4} - \frac{e^{4x}}{16} + c_1 \)
Substitute this back into the expression for \( I \):
\( I = \frac{x^2 e^{4x}}{4} - \frac{1}{2} \left[ \frac{x e^{4x}}{4} - \frac{e^{4x}}{16} \right] + c \)
\( = \frac{x^2 e^{4x}}{4} - \frac{x e^{4x}}{8} + \frac{e^{4x}}{32} + c \)
\( = e^{4x} \left[ \frac{x^2}{4} - \frac{x}{8} + \frac{1}{32} \right] + c \)
In simple words: This problem requires applying integration by parts twice because it involves a product of an algebraic term (x²) and an exponential term (e⁴ˣ). Each application reduces the power of the algebraic term until it becomes a constant.

🎯 Exam Tip: When integrating products of polynomials and exponentials, apply integration by parts repeatedly until the polynomial term becomes a constant, simplifying the integral. Be careful with calculations in each step.

Question 3. \( \int x^2 e^{3x} \, dx \)
Answer: Solution:
Let \( I = \int x^2 e^{3x} \, dx \)
Using Integration by Parts: \( \int u \, dv = uv - \int v \, du \)
Let \( u = x^2 \implies du = 2x \, dx \)
Let \( dv = e^{3x} \, dx \implies v = \int e^{3x} \, dx = \frac{e^{3x}}{3} \)
\( I = x^2 \left( \frac{e^{3x}}{3} \right) - \int \left( \frac{e^{3x}}{3} \right) (2x) \, dx \)
\( = \frac{x^2 e^{3x}}{3} - \frac{2}{3} \int x e^{3x} \, dx \)
Now, we integrate \( \int x e^{3x} \, dx \) again by parts.
Let \( u = x \implies du = 1 \, dx \)
Let \( dv = e^{3x} \, dx \implies v = \frac{e^{3x}}{3} \)
So, \( \int x e^{3x} \, dx = x \left( \frac{e^{3x}}{3} \right) - \int \left( \frac{e^{3x}}{3} \right) (1) \, dx \)
\( = \frac{x e^{3x}}{3} - \frac{1}{3} \int e^{3x} \, dx \)
\( = \frac{x e^{3x}}{3} - \frac{1}{3} \left( \frac{e^{3x}}{3} \right) + c_1 \)
\( = \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} + c_1 \)
Substitute this back into the expression for \( I \):
\( I = \frac{x^2 e^{3x}}{3} - \frac{2}{3} \left[ \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} \right] + c \)
\( I = \frac{x^2 e^{3x}}{3} - \frac{2x e^{3x}}{9} + \frac{2e^{3x}}{27} + c \)
In simple words: This problem is similar to Question 2, involving the integration of x² times e³ˣ. It requires two rounds of integration by parts to fully simplify the algebraic term.

🎯 Exam Tip: Be careful with the coefficients and signs when applying integration by parts multiple times. Organize your work clearly to avoid common errors and ensure accuracy.

Question 4. \( \int x^3 e^{x^2} \, dx \)
Answer: Solution:
Let \( I = \int x^3 e^{x^2} \, dx \)
We can rewrite \( x^3 \) as \( x^2 \cdot x \):
\( I = \int x^2 e^{x^2} \cdot x \, dx \)
Now, use substitution. Let \( t = x^2 \).

\( \implies \frac{dt}{dx} = 2x \)

\( \implies dt = 2x \, dx \)

\( \implies x \, dx = \frac{dt}{2} \)
Substitute these into the integral:
\( I = \int t e^t \cdot \frac{dt}{2} \)
\( I = \frac{1}{2} \int t e^t \, dt \)
Now, use Integration by Parts for \( \int t e^t \, dt \).
Let \( u = t \implies du = 1 \, dt \)
Let \( dv = e^t \, dt \implies v = \int e^t \, dt = e^t \)
\( \int t e^t \, dt = t e^t - \int e^t (1) \, dt \)
\( = t e^t - e^t + c_1 \)
Now substitute this back into the expression for \( I \):
\( I = \frac{1}{2} [t e^t - e^t] + c \)
\( I = \frac{1}{2} (t-1) e^t + c \)
Finally, substitute back \( t = x^2 \):
\( I = \frac{1}{2} (x^2-1) e^{x^2} + c \)
In simple words: This integral initially looks complex but can be simplified using a substitution. By letting \(t = x^2\), the integral transforms into a simpler form that can then be solved with a single application of integration by parts.

🎯 Exam Tip: Look for opportunities to use substitution (like \(u=x^2\)) before applying integration by parts, especially when the exponent of an exponential term is a function of x, as it can significantly simplify the problem.

Question 5. \( \int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx \)
Answer: Solution:
Let \( I = \int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx \)
This integral is in the special form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Let \( f(x) = \frac{1}{x} \).
Then, find the derivative of \( f(x) \):
\( f'(x) = \frac{d}{dx} (x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2} \)
We see that the integrand is exactly \( e^x [f(x) + f'(x)] \).

\( \implies I = e^x f(x) + c \)
\( I = e^x \left( \frac{1}{x} \right) + c \)
\( I = \frac{e^x}{x} + c \)
In simple words: This problem utilizes a special integration formula for integrals of the form \( \int e^x [f(x) + f'(x)] \, dx \), which simplifies directly to \( e^x f(x) + C \). Here, \( f(x) = 1/x \) and its derivative \( f'(x) = -1/x^2 \).

🎯 Exam Tip: Recognize the special form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \). This saves significant time and effort compared to direct integration by parts and is a common pattern in exams.

Question 6. \( \int e^x \frac{x}{(x+1)^2} \, dx \)
Answer: Solution:
Let \( I = \int e^x \frac{x}{(x+1)^2} \, dx \)
To use the form \( \int e^x [f(x) + f'(x)] \, dx \), we manipulate the fraction:
\( \frac{x}{(x+1)^2} = \frac{(x+1) - 1}{(x+1)^2} \)
\( = \frac{x+1}{(x+1)^2} - \frac{1}{(x+1)^2} \)
\( = \frac{1}{x+1} - \frac{1}{(x+1)^2} \)
So, the integral becomes:
\( I = \int e^x \left[ \frac{1}{x+1} - \frac{1}{(x+1)^2} \right] \, dx \)
Let \( f(x) = \frac{1}{x+1} \).
Then, find the derivative of \( f(x) \):
\( f(x) = (x+1)^{-1} \)
\( f'(x) = -1(x+1)^{-2} \cdot \frac{d}{dx}(x+1) \)
\( = - \frac{1}{(x+1)^2} \cdot (1) \)
\( = - \frac{1}{(x+1)^2} \)
We see that the integrand is \( e^x [f(x) + f'(x)] \).

\( \implies I = e^x f(x) + c \)
\( I = e^x \left( \frac{1}{x+1} \right) + c \)
\( I = \frac{e^x}{x+1} + c \)
In simple words: To solve this, we manipulate the fraction inside the integral to fit the special form \( \int e^x [f(x) + f'(x)] \, dx \). By rewriting \( x/(x+1)^2 \) as \( ( (x+1)-1 ) / (x+1)^2 \), we can identify \( f(x) = 1/(x+1) \) and its derivative \( f'(x) = -1/(x+1)^2 \).

🎯 Exam Tip: When you see \( e^x \) multiplied by a complex rational function, try to algebraically transform the rational function into the form \( f(x) + f'(x) \) to apply the direct formula. This is a crucial simplification technique.

Question 7. \( \int e^x \frac{x-1}{(x+1)^3} \, dx \)
Answer: Solution:
Let \( I = \int e^x \frac{x-1}{(x+1)^3} \, dx \)
Manipulate the numerator to match the denominator structure:
\( \frac{x-1}{(x+1)^3} = \frac{(x+1) - 2}{(x+1)^3} \)
\( = \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \)
\( = \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \)
So, the integral becomes:
\( I = \int e^x \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] \, dx \)
Let \( f(x) = \frac{1}{(x+1)^2} \).
Then, find the derivative of \( f(x) \):
\( f(x) = (x+1)^{-2} \)
\( f'(x) = -2(x+1)^{-3} \cdot \frac{d}{dx}(x+1) \)
\( = -2(x+1)^{-3} \cdot (1) \)
\( = - \frac{2}{(x+1)^3} \)
We see that the integrand is \( e^x [f(x) + f'(x)] \).

\( \implies I = e^x f(x) + c \)
\( I = e^x \left( \frac{1}{(x+1)^2} \right) + c \)
\( I = \frac{e^x}{(x+1)^2} + c \)
In simple words: Similar to Question 6, this problem requires rewriting the fraction \( (x-1)/(x+1)^3 \) as \( ( (x+1)-2 ) / (x+1)^3 \). This manipulation helps in identifying the function \( f(x) = 1/(x+1)^2 \) and its derivative, allowing the use of the special integration formula.

🎯 Exam Tip: Practice algebraic manipulation of rational functions to quickly identify \( f(x) \) and \( f'(x) \) for integrals involving \( e^x \) and fractional terms. This skill is vital for solving such problems efficiently.

Question 8. \( \int e^x \left[ (\log x)^2 + \frac{2 \log x}{x} \right] \, dx \)
Answer: Solution:
Let \( I = \int e^x \left[ (\log x)^2 + \frac{2 \log x}{x} \right] \, dx \)
This integral is in the special form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Let \( f(x) = (\log x)^2 \).
Then, find the derivative of \( f(x) \):
Using the chain rule, \( \frac{d}{dx} (g(x))^n = n (g(x))^{n-1} g'(x) \).
\( f'(x) = 2 (\log x)^{2-1} \cdot \frac{d}{dx} (\log x) \)
\( = 2 (\log x) \cdot \left( \frac{1}{x} \right) \)
\( = \frac{2 \log x}{x} \)
We see that the integrand is exactly \( e^x [f(x) + f'(x)] \).

\( \implies I = e^x f(x) + c \)
\( I = e^x (\log x)^2 + c \)
In simple words: This integral directly matches the special form \( \int e^x [f(x) + f'(x)] \, dx \). By setting \( f(x) = (\log x)^2 \), its derivative \( f'(x) \) is found to be \( 2 \log x / x \), which is exactly the second term in the integrand.

🎯 Exam Tip: This question is a straightforward application of the \( \int e^x [f(x) + f'(x)] \, dx \) formula. Quickly identify if one part is the derivative of the other within the bracket, especially with composite functions like \( (\log x)^n \).

Question 9. \( \int \left[ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right] \, dx \)
Answer: Solution:
Let \( I = \int \left[ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right] \, dx \)
This integral involves \( \log x \), so we can try a substitution.
Let \( t = \log x \).

\( \implies x = e^t \)

\( \implies \frac{dx}{dt} = e^t \)

\( \implies dx = e^t \, dt \)
Substitute these into the integral:
\( I = \int \left[ \frac{1}{t} - \frac{1}{t^2} \right] e^t \, dt \)
This is now in the special form \( \int e^t [f(t) + f'(t)] \, dt \).
Let \( f(t) = \frac{1}{t} \).
Then, find the derivative of \( f(t) \):
\( f'(t) = \frac{d}{dt} (t^{-1}) = -1 \cdot t^{-2} = -\frac{1}{t^2} \)
We see that the integrand is exactly \( e^t [f(t) + f'(t)] \).

\( \implies I = e^t f(t) + c \)
\( I = e^t \left( \frac{1}{t} \right) + c \)
Finally, substitute back \( t = \log x \) and \( e^t = x \):
\( I = x \cdot \frac{1}{\log x} + c \)
\( I = \frac{x}{\log x} + c \)
In simple words: This problem starts with an integral that doesn't immediately look like the special \( \int e^x [f(x) + f'(x)] \, dx \) form. However, a substitution of \( t = \log x \) transforms it into exactly that form, making it solvable.

🎯 Exam Tip: When you encounter integrals with \( \log x \) in the denominator, consider the substitution \( t = \log x \). This often simplifies the integral into a recognizable \( \int e^t [f(t) + f'(t)] \, dt \) form.

Question 10. \( \int \frac{\log x}{(1+\log x)^2} \, dx \)
Answer: Solution:
Let \( I = \int \frac{\log x}{(1+\log x)^2} \, dx \)
Similar to Question 9, let's use the substitution \( t = \log x \).

\( \implies x = e^t \)

\( \implies dx = e^t \, dt \)
Substitute these into the integral:
\( I = \int \frac{t}{(1+t)^2} e^t \, dt \)
Now, manipulate the fraction \( \frac{t}{(1+t)^2} \) to fit the \( f(t) + f'(t) \) form:
\( \frac{t}{(1+t)^2} = \frac{(1+t) - 1}{(1+t)^2} \)
\( = \frac{1+t}{(1+t)^2} - \frac{1}{(1+t)^2} \)
\( = \frac{1}{1+t} - \frac{1}{(1+t)^2} \)
So, the integral becomes:
\( I = \int e^t \left[ \frac{1}{1+t} - \frac{1}{(1+t)^2} \right] \, dt \)
Let \( f(t) = \frac{1}{1+t} \).
Then, find the derivative of \( f(t) \):
\( f(t) = (1+t)^{-1} \)
\( f'(t) = -1(1+t)^{-2} \cdot \frac{d}{dt}(1+t) \)
\( = - \frac{1}{(1+t)^2} \cdot (1) \)
\( = - \frac{1}{(1+t)^2} \)
We see that the integrand is exactly \( e^t [f(t) + f'(t)] \).

\( \implies I = e^t f(t) + c \)
\( I = e^t \left( \frac{1}{1+t} \right) + c \)
Finally, substitute back \( t = \log x \) and \( e^t = x \):
\( I = x \cdot \frac{1}{1+\log x} + c \)
\( I = \frac{x}{1+\log x} + c \)
In simple words: This problem combines substitution and algebraic manipulation to fit the special integration formula. First, substitute \( t = \log x \) to simplify the integrand. Then, rewrite the resulting rational expression in terms of \( t \) as \( f(t) + f'(t) \) before applying the formula.

🎯 Exam Tip: Be systematic: first apply substitution to eliminate \( \log x \), then perform algebraic rearrangement of the rational function, and finally apply the \( \int e^x [f(x) + f'(x)] \, dx \) rule. This multi-step approach is key.