Let \(2 x-3=A \frac{d}{d x}\left[x^2+4 x-12\right]+B\)
Then 2x - 3 = A(2x + 4) + B
Equating corresponding terms'on both sides,
2A = 2 ........(1)
4A + B = -3 .........(2)
from (1), A = 1
Putting this in (2)
4(1) + B = -3
\(\therefore\) B = -7
\(\therefore\) 2x - 3 = 1(2x + 4) - 7
\(\therefore \frac{2 x-3}{x^2+4 x-12}=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{x^2+4 x-12}\)
\(=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{\left(x^2+4 x+2^2\right)-2^2-12} \)
\(=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{(x+2)^2-16}\)
\(=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{(x+2)^2-4^2}\)
\(\therefore \int \frac{2 x-3}{x^2+4 x-12} d x=\int \frac{2 x+4}{x^2+4 x-12} d x -7 \int \frac{1}{(x+2)^2-4^2} d x\)
\(=\log \left|x^2+4 x-12\right|-7 \times \frac{1}{2(4)} \log \left|\frac{(x+2)-4}{(x+2)+4}\right|+c \)
\(=\log \left|x^2+4 x-12\right|-\frac{7}{8} \log \left|\frac{x-2}{x+6}\right|+c\)

Let \(\frac{3 x-9}{(x-1)(x+2)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C x+D}{x^2+1}\)
Multiplying both sides by \((x-1)(x+2)\left(x^2+1\right)\)
\( 3 x-9= \Lambda(x+2)\left(x^2+1\right)+B(x-1)\left(x^2+1\right) +(C x+D)(x-1)(x+2) .........(1)\)
Putting x = 1 in (1)
3(1) - 9 = A(1 + 2)(1² + 1) + 0 + 0
-6 = A(3(2)
-6 = 6A ⇒ A = -1
Putting x = -2 in (1)
\( 3(-2)-9 =0+B(-2-1)(4+1) \)
\(-15 =-15\ B \Rightarrow B=1\)
Equating cocfficient of x² both sides in (1)
0 =  A + B + C
0 = -1 + 1 + C
0 = C ⇒ C = 0
Equating constant term both sides in (1)
-9 = 2A - B - 2D
-9 = -2 - 1 - 2D
2D = 9 - 3 = 6 ⇒ D = 3
\(\therefore \frac{3 x-9}{(x-1)(x+2)\left(x^2+1\right)}=\frac{-1}{x-1}+\frac{1}{x+2}+\frac{0 x+3}{x^2+1}\)
\(=\frac{1}{x+2}-\frac{1}{x-1}+\frac{3}{x^2+1}\)
\( \therefore \int \frac{3 x-9}{(x-1)(x+2)\left(x^2+1\right)} d x =\int \frac{1}{x+2} d x-\int \frac{1}{x-1} d x+3 \int \frac{1}{x^2+1} d x \)
\(=\log |x+2|-\log |x-1|+3 \tan ^{-1} x+c\)
\(=\log \left|\frac{x+2}{x-1}\right|+3 \tan ^{-1} x+c\)

Let \(\frac{x+1}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}\)
Multiplying both sides by (x + 2)(x + 3)
x + 1 =  A(x + 3) + B(x + 2) ............(1)
Putting x = -2 in (1)
-2 + 1 = A(-2 + 3) + 0
-1 = A(1) ⇒ A = -1
Putting x = -3 in (1)
-3 +1 = 0 + B(-3 + 2)
-2 = B(-1) ⇒ B = 2
\(\therefore \frac{x+1}{(x+2)(x+3)}=\frac{-1}{x+2}+\frac{2}{x+3}\)
\(\therefore \int \frac{x+1}{(x+2)(x+3)} d x=\int\left[\frac{-1}{x+2}+\frac{2}{x+3}\right] d x\)
\( =-\int \frac{1}{x+2} d x+2 \int \frac{1}{x+3} d x\)
\(=-\log |x+2|+2 \log |x+3|+c \)
\(=2 \log |x+3|-\log |x+2|+c \)

\(\int {x+3\over (x+2)^2(x+1)}dx=\int{-2\over x+2}dx-\int{1\over (x+2)^2}dx+\int{2\over x+1}dx\)
\(=-2\int{1\over x+2}dx-\int{1\over (x+2)^2}dx+2\int{1\over x+1}dx\)
\(=-2log|x+2|-\int{ (x+2)^{-2}}dx+2log{|x+1|}+c\)
\(=-2log|x+2|+{1\over x+2}+2log{|x+1|}+c\)
 

\( \int \frac{1}{\sqrt{x+1}+\sqrt{x}} d x=\int \frac{1}{\sqrt{x+1}+\sqrt{x}}\left[\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}\right] d x\)
\(=\int \frac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x+1}^2\right)-(\sqrt{x})^2} d x\)
\(=\int \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} d x=\int(\sqrt{x+1}-\sqrt{x}) d x\)
\(=\int \sqrt{x+1} d x-\int \sqrt{x} d x=\int(x+1)^{\frac{1}{2}} d x-\int x^{\frac{1}{2}} d x\)
\( =\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{X^{\frac{3}{2}}}{\frac{3}{2}}+C\)
\(=\frac{2}{3}\left[(x+1)^{\frac{3}{2}}-X^{\frac{3}{2}}\right]+C . \)

Let a be the acceleration, v be the velocity of the car, and s be the distance.
Stated in calculus terminology, velocity, v = \({ds\over dt}\), is the rate of change of position with time, and acceleration, a = \({dv\over dt}\), is rate of change of velocity with time.
The acceleration to be negative because if you take the direction of movement to be positive, then for a bike that is slowing down, its acceleration vector will be oriented in the opposite direction of its motion (retardation).
Given that the retardation of the car is 8 meter/second².
Therefore, a = \({dv\over dt}=-8 \) meter/second²
Therefore, \(v=\int a\ dt=\int -8dt=-8t+c_1\)
\(v=-8t+c_1.\)
When the brakes are applied,
\(t=0, and \ v=24m/s.\)
So, 24 = -8(0) + c₁ \(\Rightarrow c_1=24\)
Therefore, v = -8t + 24.
That is, \({ds\over dt }=-8t+24.\)
It is required to find the distance, not the velocity, so need more integration in order.
\(s=\int vdt=\int (-8t+24)dt\)
\(s=-4t^2+24t+c_2\)
To determine c₂, the stopping distance s is measured from where, and when, the brakes are applied so that at t = 0, s = 0.
\(s=-4t^2+24t+c_2\Rightarrow 0=-4(0)^2+24(0)+c_2 \Rightarrow c_2=0\)
\(s=-4t^2+24t\)
The stopping distance s could be evaluated if we knew the braking time. The time can be determined from the speed statement.
The bike stops when v = 0, \(\Rightarrow v=-8t+24 \Rightarrow 0=-8t+24\Rightarrow t=3.\)
When t = 3 , we get
\(s=-4t^2+24t\Rightarrow s=-4(3)^2+24(3)\)
s = 36 metres < 40 metres
The bike stops at a distance 4 metres to the barrier.

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.

In calculus terminology, velocity v = \({ds \over dt}\) is rate of change of position with time, where s is the distance. The velocity of the train is given by
\(v(t)=20t+50\)
Therefore, \({ds \over dt}=20t+50\)
To find the distance function s one has to integrate the derivative function.
That is, \(s=\int (20t+50)dt\)
\(s=10t^2+50t+c\)
The distance covered by the train is zero when time is zero. Let us use this initial condition s = 0 at t = 0 to determine the value c of the constant of integration.
\(\Rightarrow \ s=10t^2+50t+c \Rightarrow c=0\)
Therefore, \(s=10t^2+50t\)
The distance covered by the train in 2 hours (5 pm - 3 pm) is given by substituting t = 2 in the above equation, we get
s = 10(2)² + 50(2) = 140 km.

Given that f'(x) = \({d\over dx}(f(x))=3x^2-4x+5\)
Integrating on both sides with respect to x, we get
\(\int f'(x)dx=\int (3x^2-4x+5)dx\)
\(f(x)=x^3-2x^2+5x+c\)
To determine the constant of integration c, we have to apply the given information f(1) = 3
\(f(1)=3\Rightarrow 3=(1)^3-2(1)^2+5(1)+c \Rightarrow c=-1\)
Thus \(f(x)=x^3-2x^2+5x-1\).