Rd Sharma XII Vol 1 2018 Solutions for Class 12 Science Math Chapter 19 Indefinite Integrals are provided here with simple step-by-step explanations. These solutions for Indefinite Integrals are extremely popular among Class 12 Science students for Math Indefinite Integrals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2018 Book of Class 12 Science Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2018 Solutions. All Rd Sharma XII Vol 1 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

Comparing the Coefficients of like powers of x we get

#### Question 2:

Comparing Coefficients of like powers of x

#### Question 3:

Comparing Coefficients of like powers of x

$2\mathrm{A}=1\phantom{\rule{0ex}{0ex}}\mathrm{A}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}2\mathrm{A}+\mathrm{B}=-3\phantom{\rule{0ex}{0ex}}2×\frac{1}{2}+\mathrm{B}=-3\phantom{\rule{0ex}{0ex}}\mathrm{B}=-4$

#### Question 4:

Comparing Coefficients of like powers of x

#### Question 5:

Comparing the Coefficients of like powers of x

#### Question 6:

Comparing the Coefficients of like powers of x

#### Question 7:

Comparing the Coefficients of like powers of x

#### Question 8:

Comparing the Coefficients of like powers of x

#### Question 10:

Comparing the Coefficients of like powers of t

#### Question 11:

Comparing the Coefficients of like powers of x

#### Question 12:

Evaluate the following integrals:

$\int \frac{5x-2}{1+2x+3{x}^{2}}\mathrm{d}x$

#### Question 13:

$\int \frac{x+5}{3{x}^{2}+13x-10}dx$

$I=\int \frac{x+5}{3{x}^{2}+13x-10}dx\phantom{\rule{0ex}{0ex}}=\int \frac{x+5}{3{x}^{2}+15x-2x-10}dx\phantom{\rule{0ex}{0ex}}=\int \frac{x+5}{3x\left(x+5\right)-2\left(x+5\right)}dx\phantom{\rule{0ex}{0ex}}=\int \frac{x+5}{\left(3x-2\right)\left(x+5\right)}dx$

#### Question 14:

$\phantom{\rule{0ex}{0ex}}$
Using partial fraction, we get

$\frac{\left(3t-2\right)}{\left(t-3\right)\left(t-4\right)}=\frac{A}{\left(t-3\right)}+\frac{B}{\left(t-4\right)}=\frac{A\left(t-4\right)+B\left(t-3\right)}{\left(t-3\right)\left(t-4\right)}\phantom{\rule{0ex}{0ex}}⇒3t-2=\left(A+B\right)t-4A-3B$

Comparing coefficients, we get

A = $-$7 and B = 10

So, $I=-7\int \frac{1}{\left(t-3\right)}dt+10\int \frac{1}{\left(t-4\right)}dt$

#### Question 15:

$\int \frac{x+7}{3{x}^{2}+25x+28}dx$

$I=\int \frac{x+7}{3{x}^{2}+25x+28}dx\phantom{\rule{0ex}{0ex}}=\int \frac{x+7}{3{x}^{2}+21x+4x+28}dx\phantom{\rule{0ex}{0ex}}=\int \frac{x+7}{3x\left(x+7\right)+4\left(x+7\right)}dx\phantom{\rule{0ex}{0ex}}=\int \frac{x+7}{\left(3x+4\right)\left(x+7\right)}dx$

$=\int \frac{1}{\left(3x+4\right)}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{ln}\left|3x+4\right|+c$

#### Question 16:

Evaluate the following integrals:
$\int \frac{{x}^{3}}{{x}^{4}+{x}^{2}+1}dx$

$=\frac{1}{4}\int \left[\frac{\left(2t+1\right)}{\left({t}^{2}+t+1\right)}-\frac{1}{\left({t}^{2}+t+1\right)}\right]dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[\mathrm{log}\left|{t}^{2}+t+1\right|-\int \frac{1}{\left({t}^{2}+t+\frac{1}{4}+\frac{3}{4}\right)}dt\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[\mathrm{log}\left|{t}^{2}+t+1\right|-\int \frac{1}{{\left(t+\frac{1}{2}\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}}dt\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[\mathrm{log}\left|{t}^{2}+t+1\right|-\frac{2}{\sqrt{3}}\mathrm{tan}\frac{\left(t+\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}\right]+c\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[\mathrm{log}\left|{t}^{2}+t+1\right|-\frac{2}{\sqrt{3}}\mathrm{tan}\left(\frac{2t+1}{\sqrt{3}}\right)\right]+c$

$=\frac{1}{4}\left[\mathrm{log}\left|{x}^{4}+{x}^{2}+1\right|-\frac{2}{\sqrt{3}}\mathrm{tan}\left(\frac{2{x}^{2}+1}{\sqrt{3}}\right)\right]+c$

#### Question 17:

Evaluate the following integrals:
$\int \frac{{x}^{3}-3x}{{x}^{4}+2{x}^{2}-4}dx$

$I=\int \frac{{x}^{3}-3x}{{x}^{4}+2{x}^{2}-4}dx$

Let ${x}^{2}=t$, or, $2xdx=dt$

$⇒I=\frac{1}{2}\int \frac{\left(t-3\right)}{{t}^{2}+2t-4}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\int \frac{2t-6}{{t}^{2}+2t-4}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\int \frac{2t+2-8}{{t}^{2}+2t-4}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\int \left(\frac{2t+2}{{t}^{2}+2t-4}-\frac{8}{{t}^{2}+2t-4}\right)dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left(\int \frac{2t+2}{{t}^{2}+2t-4}dt-\int \frac{8}{{t}^{2}+2t-4}dt\right)$

Now,

Let ${t}^{2}+2t-4=u$

Now,

#### Question 3:

Comparing coefficients of like terms

#### Question 18:

Evaluate the following integrals:

$\int \frac{x+2}{\sqrt{{x}^{2}+2x+3}}\mathrm{d}x$

#### Question 13:

$=-\frac{1}{2}\mathrm{log}\left|\frac{\mathrm{tan}\frac{x}{2}-2-\sqrt{3}}{\mathrm{tan}\frac{x}{2}+2-\sqrt{3}}\right|+C\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{log}\left|\frac{\mathrm{tan}\frac{x}{2}+2-\sqrt{3}}{\mathrm{tan}\frac{x}{2}+2-\sqrt{3}}\right|+C$

#### Question 3:

Comparing the coefficients of like terms

Multiplying eq (1) by 2 and adding it to eq (2) we get ,

$⇒4A-2B+A+2B=8+2\phantom{\rule{0ex}{0ex}}⇒5A=10\phantom{\rule{0ex}{0ex}}⇒A=2$

Putting value of A = 2 in  eq (1)

#### Question 4:

Comparing coefficients of like terms

Multiplying eq (1) by p and eq (2) by q and then adding

$⇒A{p}^{2}+Bpq=p\phantom{\rule{0ex}{0ex}}⇒A{q}^{2}-Bpq=0\phantom{\rule{0ex}{0ex}}⇒A=\frac{p}{{p}^{2}{q}^{2}}$

Putting value of A in eq (1)

#### Question 5:

Comparing coefficients of like terms

Multiplying eq (3) by 2 and then adding to eq (2)

4A + 2B + A – 2B = 10
$⇒$A = 2

Putting value of A in eq (2) and eq (4) we get,
B = 1& C = 0

#### Question 6:

By comparing the coefficients of like terms we get,

Multiplying eq (2) by 3 and eq (3) by 4 and then adding,

Thus, substituting the values of A,B and C in eq (1) we get ,

#### Question 7:

Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get

#### Question 8:

Multiplying equation (2) by 3 and equation (3) by 4 ,then by adding them we get

#### Question 9:

Solving (1) and (2), we get

#### Question 10:

Solving eq (2) and  eq (3) we get,
A = 2, B = 1
Thus, by substituting the values of A and B in eq (1) we get ,

#### Question 11:

By solving eq (2) and eq (3) we get,

#### Question 27:

Evaluate the following integrals:

#### Question 28:

Evaluate the following integrals:

$\int \frac{\mathrm{log}x}{{\left(x+1\right)}^{2}}\mathrm{d}x$

#### Question 32:

$\int$ x tan2 x dx
= ​∫ x (sec2 x – 1) dx

#### Question 35:

$\int$ sin–1 (3x – 4x3)dx
Let x = sin θ
dx = cos​ θ.dθ
& θ = sin–1 x
$\int$ sin–1 (3x – 4x3)dx =$\int$ sin–1 (3 sin ​θ – 4 sin3 ​θ) . cos ​θ d​θ
= ∫ sin–1 (sin 3​θ) . cos ​θ d​θ

#### Question 51:

Let  I =$\int$ (tan–1 x2) x dx
Putting x2 = t
⇒​ 2x dx = dt

#### Question 53:

Note: The answer in indefinite integration may vary depending on the integral constant.

#### Question 54:

Let I =
sin (3A) = 3 sin A – 4 sin3 A

#### Question 55:

Note: The final answer in indefinite integration may vary based on the integration constant.

Let I=

#### Question 1:

$\int \left(3x\sqrt{x}+4\sqrt{x}+5\right)dx$

#### Question 2:

$\int \left({2}^{x}+\frac{5}{x}-\frac{1}{{x}^{1/3}}\right)dx$

#### Question 9:

$\int \left({x}^{e}+{e}^{x}+{e}^{e}\right)dx\phantom{\rule{0ex}{0ex}}=\int {x}^{e}dx+\int {e}^{x}dx+{e}^{e}\int 1dx\phantom{\rule{0ex}{0ex}}=\frac{{x}^{e+1}}{e+1}+{e}^{x}+x·{e}^{e}+C$

#### Question 24:

Evaluate the following integrals:

$\int {\mathrm{e}}^{2x}\left(\frac{1-\mathrm{sin}2\mathrm{x}}{1-\mathrm{cos}2\mathrm{x}}\right)\mathrm{d}x$