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#### 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:

$\int \frac{3x+5}{{x}^{2}+3x-18}dx$

$I=\int \frac{3x+5}{{x}^{2}+3x-18}dx$

$⇒I=3\int \frac{x+\frac{5}{3}}{{x}^{2}+3x-18}dx$

$⇒I=\frac{3}{2}\int \frac{2x+\frac{10}{3}}{{x}^{2}+3x-18}dx$

$⇒I=\frac{3}{2}\int \frac{2x+3+\frac{1}{3}}{{x}^{2}+3x-18}dx$

$⇒I=\frac{3}{2}\int \frac{2x+3}{{x}^{2}+3x-18}dx+\frac{1}{2}\int \frac{1}{{x}^{2}+3x-18}dx$

$⇒I=\frac{3}{2}\mathrm{log}\left|{x}^{2}+3x-18\right|+\frac{1}{2}\int \frac{1}{{x}^{2}+3x+\frac{9}{4}-18-\frac{9}{4}}dx$         $\left(\begin{array}{c}\int \frac{f\text{'}\left(x\right)dx}{f\left(x\right)}=\mathrm{log}\left|f\left(x\right)\right|+C\\ f\left(x\right)={x}^{2}+3x-18⇒f\text{'}\left(x\right)=2x=3\end{array}\right)$

$⇒I=\frac{3}{2}\mathrm{log}\left|{x}^{2}+3x-18\right|+\frac{1}{2}\int \frac{1}{{\left(x+\frac{3}{2}\right)}^{2}-\frac{81}{4}}dx$

$⇒I=\frac{3}{2}\mathrm{log}\left|{x}^{2}+3x-18\right|+\frac{1}{2}\int \frac{1}{{\left(x+\frac{3}{2}\right)}^{2}-{\left(\frac{9}{2}\right)}^{2}}dx$

$⇒I=\frac{3}{2}\mathrm{log}\left|{x}^{2}+3x-18\right|+\frac{1}{2}×\frac{1}{2×\frac{9}{2}}\mathrm{log}\left|\frac{\left(x+\frac{3}{2}\right)-\frac{9}{2}}{\left(x+\frac{3}{2}\right)+\frac{9}{2}}\right|+C$                $\left(\int \frac{dx}{{x}^{2}-{a}^{2}}=\frac{1}{2a}\mathrm{log}\left|\frac{x-a}{x+a}\right|+C\right)$

$⇒I=\frac{3}{2}\mathrm{log}\left|{x}^{2}+3x-18\right|+\frac{1}{18}\mathrm{log}\left|\frac{x-3}{x+6}\right|+C$

#### Question 17:

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 18:

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

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#### 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
= â€‹∫ x (sec2 x – 1) dx

#### Question 35:

$\int$ sin–1 (3x – 4x3)dx
Let x = sin θ
dx = cosâ€‹ θ.dâ€‹θ
& θ = sin–1 x
$\int$ sin–1 (3x – 4x3)dx =$\int$ sin–1 (3 sin â€‹θ – 4 sin3 â€‹θ) . cos â€‹θ dâ€‹θ
= ∫ sin–1 (sin 3â€‹θ) . cos â€‹θ dâ€‹θ

#### Question 51:

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

#### Question 53:

$\int {\mathrm{sin}}^{-1}2xdx$

$=\int {\mathrm{sin}}^{-1}2x×1dx$

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

$=x{\mathrm{sin}}^{-1}2x-\int \frac{2}{\sqrt{1-4{x}^{2}}}×xdx$

$=x{\mathrm{sin}}^{-1}2x-\int \frac{2x}{\sqrt{1-4{x}^{2}}}dx$

$=x{\mathrm{sin}}^{-1}2x+\frac{1}{4}\int {\left(1-4{x}^{2}\right)}^{-\frac{1}{2}}\left(-8x\right)dx$

#### Question 54:

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

#### Question 55:

Let I =
sin (3A) = 3 sin A – 4 sin3 A

#### Question 56:

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:

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