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RD Sharma Class 12 Exercise 18.15 Indefinite Integrals Solutions Maths - Download PDF Free Online
It is a well-known fact that RD Sharma class 12th exercise 18.15 is a highly recommended NCERT solution. Students and teachers from across the Indian subcontinent have expressed their trust in RD Sharma Solutions, which has become a popular name in the education industry. The RD Sharma class 12 chapter 18 exercise 18.15 is especially favoured by students as it contains answers to a challenging section from the maths book.
This Story also Contains
- RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
- Indefinite Integrals Excercise: 18.15
- RD Sharma Chapter wise Solutions
The RD Sharma class 12 solutions Indefinite Integrals 18.15 will cover the chapter Indefinite Integrals which is complex and pretty lengthy. The chapter will test stunning on the concepts of Integrals, Graphs of indefinite integrals, Indefinite integrals of common functions, solving the quadratic equation etc. The exercise 18.15 part has a total of 6 questions. The RD Sharma class 12th exercise 18.15 will provide students with easy answers to all 6 of them.
RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Chapter 18 - Indefinite Integrals - Ex-18.1
Indefinite Integrals Excercise: 18.15
Indefinite Integrals Exercise 18.15 Question 1
Answer:$\frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{4 x^{2}+12 x+5} d x$
Solution:
Let $I=\int \frac{1}{4 x^{2}+12 x+5} d x=\frac{1}{4} \int \frac{1}{x^{2}+\frac{12 x}{4}+\frac{5}{4}} d x$
$\begin{aligned} &=\frac{1}{4} \int \frac{1}{x^{2}+3 x+\frac{5}{4}} d x=\frac{1}{4} \int \frac{d x}{x^{2}+2 \cdot x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{4}} \\\\ &=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}+\frac{5}{4}} d x=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9-5}{4}\right)} d x \end{aligned}$
Put $x+\frac{3}{2}=t \Rightarrow d x=d t$
Then $I=\frac{1}{4} \int \frac{1}{t^{2}-1^{2}} d t$
$=\frac{1}{4} \cdot \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\right|+C$ $\quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$
$=\frac{1}{4} \times \frac{1}{2} \log \left|\frac{x+\frac{3}{2}-1}{x+\frac{3}{2}+1}\right|+C \text { }$ $\left[\because t=x+\frac{3}{2}\right]$
$\begin{aligned} &=\frac{1}{8} \log \left|\frac{\frac{2 x+3-2}{2}}{\frac{2 x+3+2}{2}}\right|+C \\\\ &\end{aligned}$
$=\frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C$
Indefinite Integrals Exercise 18.15 Question 2
Answer:$\frac{1}{3} \tan ^{-1}\left(\frac{x-5}{3}\right)+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{x^{2}-10 x+34} d x$
Solution:
Let $I=\int \frac{1}{x^{2}-10 x+34} d x$
$\begin{aligned} &=\int \frac{1}{x^{2}-2 \cdot x \cdot 5+5^{2}-5^{2}+34} d x \\\\ &=\int \frac{1}{(x-5)^{2}-25+34} d x=\int \frac{1}{(x-5)^{2}+9} d x \\\\ &=\int \frac{1}{(x-5)^{2}+3^{2}} d x \end{aligned}$
Put $x-5=t \Rightarrow d x=d t$
Then $I=\int \frac{1}{t^{2}+3^{2}} d t$
$=\frac{1}{3} \tan ^{-1}\left(\frac{t}{3}\right)+C$ $\quad\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]$
$=\frac{1}{3} \tan ^{-1}\left(\frac{x-5}{3}\right)+C$ $\quad[\because t=x-5]$
Indefinite Integrals Exercise 18.15 Question 3
Answer:$\frac{-1}{\sqrt{5}} \log \left|\frac{2 x-1-\sqrt{5}}{2 x-1+\sqrt{5}}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{1+x-x^{2}} d x$
Solution:
Let $I=\int \frac{1}{1+x-x^{2}} d x=-\int \frac{1}{x^{2}-x-1} d x$
$\begin{aligned} &=-\int \frac{1}{x^{2}-2 \cdot x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-1} d x \\ & \end{aligned}$
$=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4}-1} d x$
$\begin{aligned} &=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1+4}{4}\right)} d x \\ & \end{aligned}$
$=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\frac{5}{4}} d x$
Put $x-\frac{1}{2}=t \Rightarrow d x=d t$
Then $I=-\int \frac{1}{t^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}} d t$
$=\frac{-1}{2 \cdot \frac{\sqrt{5}}{2}} \log \left|\frac{t-\frac{\sqrt{5}}{2}}{t+\frac{\sqrt{5}}{2}}\right|+C \text { }$ $\quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$
$=\frac{-1}{\sqrt{5}} \log \left|\frac{x-\frac{1}{2}-\frac{\sqrt{5}}{2}}{x-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|+C$ $\quad\left[\because t=x-\frac{1}{2}\right]$
$\begin{aligned} &=\frac{-1}{\sqrt{5}} \log \left|\frac{\frac{2 x-1-\sqrt{5}}{2}}{\frac{2 x-1+\sqrt{5}}{2}}\right|+C \\ & \end{aligned}$
$=\frac{-1}{\sqrt{5}} \log \left|\frac{2 x-1-\sqrt{5}}{2 x-1+\sqrt{5}}\right|+C$
Indefinite Integrals Exercise 18.15 Question 4
Answer:$\frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{2 x^{2}-x-1} d x$
Solution:
Let $I=\int \frac{1}{2 x^{2}-x-1} d x=\frac{1}{2} \int \frac{1}{x^{2}-\frac{x}{2}-\frac{1}{2}} d x$
$\begin{aligned} &=\frac{1}{2} \int \frac{1}{x^{2}-2 \cdot x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}-\frac{1}{2}} d x \\ & \end{aligned}$
$=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{1}{16}-\frac{1}{2}} d x$
$\begin{aligned} &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\left(\frac{1+8}{16}\right)} d x \\ & \end{aligned}$
$=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}} d x$
Put $x-\frac{1}{4}=t \Rightarrow d x=d t$
Then $I=\frac{1}{2} \int \frac{1}{t^{2}-\left(\frac{3}{4}\right)^{2}} d x$
$=\frac{1}{2} \cdot \frac{1}{2 \times \frac{3}{4}} \log \left|\frac{t-\frac{3}{4}}{t+\frac{3}{4}}\right|+C$ $\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$
$=\frac{1}{3} \log \left|\frac{x-\frac{1}{4}-\frac{3}{4}}{x-\frac{1}{4}+\frac{3}{4}}\right|+C \text { }$ $\quad\left[\because t=x-\frac{1}{4}\right]$
$\begin{aligned} &=\frac{1}{3} \log \left|\frac{\frac{4 x-1-3}{4}}{\frac{4 x-1+3}{4}}\right|+C \\ & \end{aligned}$
$=\frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C$
Indefinite Integrals Excercise 18.15 Question 5
Answer:$\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{1}{x^{2}+6 x+13} d x$
Solution:
Let $I=\int \frac{1}{x^{2}+6 x+13} d x$
$\begin{aligned} &=\int \frac{1}{x^{2}+2 \cdot x \cdot 3+3^{2}-3^{2}+13} d x \\ & \end{aligned}$
$=\int \frac{1}{(x+3)^{2}-9+13} d x \\$
$=\int \frac{1}{(x+3)^{2}+4} d x$
Put $x+3=t \Rightarrow d x=d t$
Then $I=\int \frac{1}{t^{2}+2^{2}} d t$
$=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C$ $\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]$
$=\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$ $\quad[\because t=x+3]$
Indefinite Integrals Excercise 18.15 Question 6
Answer:$\log \left|\frac{(x+3)^{2}}{(x+2)}\right|+C$
Hint:
To solve this problem use special integration formula
Given:
$\int \frac{(x+1)}{(x+2)(x+3)} d x$
Solution:
Let $I=\int \frac{(x+1)}{(x+2)(x+3)} d x$
Applying partial fraction
$\begin{aligned} &\frac{(x+1)}{(x+2)(x+3)}=\frac{A}{(x+2)}+\frac{B}{(x+3)} \\ & \end{aligned}$
$(x+1)=A(x+3)+B(x+2)$
Putting
$\begin{aligned} &x=-2 \\ \end{aligned}$
$(-2+1)=A(-2+3)+B(-2+2) \\$
$A=-1$
Putting
$\begin{aligned} &x=-3 \\ \end{aligned}$
$(-3+1)=A(-3+3)+B(-3+2) \\$
$-2=0+B(-1) \\$
$B=2$
$\begin{aligned} &I=\int \frac{x+1}{(x+1)(x+2)} d x=\int \frac{-1}{x+2} d x+\int \frac{2}{x+3} d x \\ \end{aligned}$
$=-\log |x+2|+2 \log |x+3| \\$
$=\log \left|\frac{(x+3)^{2}}{(x+2)}\right|+C$
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RD Sharma Chapter wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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