Let \(3 x+1=A \frac{d}{d x}\left(2 x^2-2 x+3\right)+B\)
Then 3x + 1 = A(4x - 2) + B
Equatingcorrespondingcoefficients onboth sides
4A = 3 .......(1)
-2A + B = 1 ...........(2)
From (1), \(A=\frac{3}{4}\)
Putting this in (2)
\(-2\left(\frac{3}{4}\right)+B =1 \)
\(\frac{-3}{2}+B =1 \)
\(B =1+\frac{3}{2}=\frac{5}{2} \)
\(\therefore 3 x+1 =\frac{3}{4}(4 x-2)+\frac{5}{2}\)
\(\therefore \frac{3 x+1}{2 x^2-2 x+3}=\frac{3}{4}\left(\frac{4 x-2}{2 x^2-2 x+3}\right)+\frac{5}{2}\left(\frac{1}{2 x^2-2 x+3}\right)\)
Now \(2 x^2-2 x+3=\left(2 x^2-2 x\right)+3\)
\(=2\left(x^2-x\right)+3\ Completing\ the\ square\ on\ x\)
\(=2\left(x^2-x+\left(\frac{1}{2}\right)^2\right)-2\left(\frac{1}{2}\right)^2+3\)
\(=2\left(x-\frac{1}{2}\right)^2-2\left(\frac{1}{4}\right)+3 \)
\(=2\left(x-\frac{1}{2}\right)^2-\frac{1}{2}+3\)
\(=2\left(x-\frac{1}{2}\right)^2+\frac{5}{2}\)
\(=2\left[\left(x-\frac{1}{2}\right)^2+\frac{5}{4}\right] \)
\(=2\left[\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{5}}{2}\right)^2\right]\)
\(\therefore \frac{3 x+1}{2 x^2-2 x+3}=\frac{3}{4}\left(\frac{4 x-2}{2 x^2-2 x+3}\right) +\frac{5}{4}\left(\frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{5}}{2}\right)^2}\right) \)
\(\therefore \int \frac{3 x+1}{2 x^2-2 x+3} d x = \frac{3}{4} \log \left|2 x^2-2 x+3\right| +\frac{5}{4} \times \frac{1}{(\sqrt{5} / 2)} \tan ^{-1}\left(\frac{x-\frac{1}{2}}{\sqrt{5} / 2}\right)+0 \)
\(=\frac{3}{4} \log \left|2 x^2-2 x+3\right| +\frac{\sqrt{5}}{2} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c \)

Let I = \(\int{2x+3\over \sqrt{x^2+x+1}}dx\)
\(2x+3=A{d\over dx}(x^2+x+1)+B\)
\(2x+3=A(2x+1)+B\)
Comparing the coefficients of like terms, we get
2A = 2 \(\Rightarrow \ A=1; A+B=3 \Rightarrow B=2\)
I =\(\int{(2x+1)+2\over \sqrt{x^2+x+1}}dx\)
I = \(\int{2x+1\over \sqrt{x^2+x+1}}dx+2\int{1\over \sqrt{x^2+x+1}}dx\)
\(=2\sqrt{x^2+x+1}+2\int{1\over \sqrt{(x+{1\over 2})^2}+({\sqrt{3}\over2})^2}dx\)
\(=2\sqrt{x^2+x+1}+2log|x+{1\over2}+\sqrt{(x+{1\over2})^2+({\sqrt{3}\over2})^2}|+c\)
Therefore, I = \(2\sqrt{x^2+x+1}+2log|x+{1\over2}+\sqrt{x^2+x+1}|+c\)

Let \(\frac{1}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
Multiplying both sides by \((x-1)(x+2)^2\)
\(1=A(x+2)^2+B(x-1)(x+2)+C(x-1)\)
Putting x = 1 in (1)
\(1=A(1+2)^2+0+0\)
\(1=9 A \Rightarrow A=\frac{1}{9}\)
Putting x = -2 in (1)
1 = 0 + 0 + C(-2 - 1)
\(\therefore 1=-3 C \Rightarrow C=-\frac{1}{3}\)
Equating coefticient of x² both sides in (1)
\(0=A+B \Rightarrow B=-A=-\frac{1}{9}\)
\(\therefore \frac{1}{(x-1)(x+2)^2}=\frac{1}{9(x-1)}-\frac{1}{9(x+2)}-\frac{1}{3(x+2)^2}\)
\( \therefore \int \frac{1}{(x-1)(x+2)^2} d x= \frac{1}{9} \log |x-1|- \frac{1}{9} \log |x+2|-\frac{1}{3}\left(\frac{-1}{x+2}\right)+c \)
\(=\frac{1}{9} \log \left|\frac{x-1}{x+2}\right|+\frac{1}{3(x+2)}+c\)

The rate of change of height h with respect to time t is the derivative of h with respect to t.
Therefore,\({dh\over dt}={18\over \sqrt{t}}=18t^{-{1\over2}}\)
So, to get a general expression for the height, integrating the above equation with respect to t.
\(h=\int {18t^{-{1\over 2}}}dt=18(2t^{1\over2})+c=36\sqrt{t}+c\)
Given that when t = 0, the height h = 5 cm.
\(5=0+c \Rightarrow c=5\)
\(h=36\sqrt{t}+5\).
(i) To find the height of the tree after 4 years.
When t = 4 years,
\(h=36\sqrt{t}+5\Rightarrow h=36\sqrt{4}+5=77\)
The height of the tree after 4 years is 77 cm .
(ii) When h = 149 cm
\(h=36 \sqrt{t}+5 \Rightarrow 149=36 \sqrt{t}+5\)
\(\sqrt{t}=\frac{149-5}{36}=4 \Rightarrow t=16\)
Thus after 16 years the height of the tree will be 149 cm.

The rate of change of weight with respect to height is
\({dw\over dh}=4.364 \times 10^{-5}h^{2}\)
\(w=\int 4.364 \times 10^{-5}h^2 dh\)
\(w= 4.364 \times 10^{-5}({h^3\over 3})+c\)
One can obviously understand that the weight of a person is zero when height is zero.
Let us find the value c of the constant of integration by substituting the initial condition w = 0, at h = 0, in the above equation
\(w=4.364 \times 10^{-5}({h^3\over 3})+c \Rightarrow c=0\)
The required relation between weight and height of a person is
\(w=4.364 \times 10^{-5}({h^3\over 3})\)
When the height h =150cm,
\(w=4.364 \times 10^{-5}({150^3\over 3})\)
When the height h = 150cm, the weight is w= 49kg (approximately)
Therefore, the weight of the person whose height 150cm is 49 kg.

Let I=\(\int e^x{(1-x)^2\over (1+x^2)^2}dx\)
\(=\int e^x{(1-x^2-2x)\over (1+x^2)^2}dx\)
\(=\int e^x({1\over (1+x^2)}-{2x\over (1+x^2)^2})dx\)
If f(x) = \({1\over (1+x^2)},\)then \(f '(x)=-{2x\over (1+x^2)^2}\)
Using \(\int e^x(f(x)+f ' (x))dx=e^xf(x)+c\)
\(\int e^x({1-x \over 1+x^2})^2dx=\int e^x({1\over (1+x^2)}-{2x\over (1+x^2)^2})dx=e^x{1\over (1+x^2)}+c\)

\(=\int x^2 \cos x d x\)
\(=x^2(\sin x)-2 x(-\cos x)+2(-\sin x)+c \)
\(=x^2 \sin x+2 x \cos x-2 \sin x+c\)
\(=\left(x^2-2\right) \sin x+2 x \cos x+c\)

\(=\int 27 x^2 e^{3 x} d x\)
\(=27 x^2\left(\frac{e^{3 x}}{3}\right)-54 x\left(\frac{e^{3 x}}{3 \times 3}\right)+54\left(\frac{e^{3 x}}{3 \times 3 \times 3}\right)+c \)
\( =9 x^2 e^{3x}-6 x e^{3x}+2 e^{3x}+c\)
\(=\left(9 x^2-6 x+2\right) e^{3x}+c\)

\(\int x^3 e^{-x}dx\)
Applying Bernoulli’s formula
\(\int udv =uv-u' v_1+u"v_2....\)
\(\int x^3 e^{-3} dx=(x^3)(-e^{-x})-(3x^2)(e^{-x})+(6x)(-e^{-x})-(6)(e^{-x})+c\)
\(=-x^3e^{-x}-3x^2e^{-x}-6xe^{-x}-6e^{-x}+c\)

\(\int(3 x+4) \sqrt{3 x+7} d x =\int[(3 x+7)-3](3 x+7)^{1 / 2} d x \)
\(=\int\left[(3 x+7)^{3 / 2}-3(3 x+7)^{1 / 2}\right] d x \)
\(=\frac{(3 x+7)^{5 / 2}}{5 / 2 \times 3}-3\left(\frac{(3 x+7)^{3 / 2}}{3 / 2 \times 3}\right)+c \)
\(=\frac{2}{15}(3 x+7)^{5 / 2}-\frac{2}{3}(3 x+7)^{3 / 2}+c \)