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1

Answer:

#-lnx/(1+x)+lnx-ln(1+x)+C#

Explanation:

#I=intlnx/(1+x)^2dx#

Integration by parts takes the form #intudv=uv-intvdu#. Let:

#u=lnx#
#dv=1/(1+x)^2dx#

Differentiate #u# and integrate #dv#. The integration of #dv# is best performed with the substitution #t=1+x=>dt=dx#. You should get:

#du=1/xdx#
#v=-1/(1+x)#

Then:

#I=uv-intvdu#

#I=-lnx/(1+x)-int1/x(-1/(1+x))dx#

#I=-lnx/(1+x)+int1/(x(1+x))dx#

Perform partial fraction decomposition on #1/(x(1+x))#:

#1/(x(1+x))=A/x+B/(1+x)#

Then:

#1=A(1+x)+Bx#

Letting #x=-1#:

#1=A(1-1)+B(-1)#

#B=-1#

Letting #x=0#:

#1=A(1+0)+B(0)#

#1=A#

Then:

#1/(x(1+x))=1/x-1/(1+x)#

So:

#I=-lnx/(1+x)+int1/xdx-int1/(1+x)dx#

These are simple integrals. The second can be performed with the substitution #s=1+x=>ds=dx#.

#I=-lnx/(1+x)+lnx-ln(1+x)+C#

1

Answer:

Chaucer should not be considered anti-feminist. He cast a realistic eye on the foibles of men and women of his day.

Explanation:

Chaucer notes the vanities, motivations, weaknesses and contradictions in the behaviour of the pilgrims. He also touches upon the faults in the society of his time. He saw human beings in all their fragility.
One should also note he was not calling for revolutionary change. At that period the Middle Ages, while undergoing change through increasing trade and education, had not evolved to the challenging spirit that came with the Renaissance.

1

Answer:

"Animal Farm" by George Orwell

Explanation:

On the surface the story seems to be a children's fable about farm animals revolting against neglect by their drunken and careless owner. The mature reader soon sees that it is a brilliant and well-informed critique of what happens in society when there is a revolution. George Orwell was well aware of the Russian Revolution, the rise of fascism after World War I and, in fact, personally experienced events during the Spanish Civil War.

1

Answer:

It was Nazi Germany's last attempt to win the war. It was the largest battle on the Western Front.

Explanation:

The Battle of the Bulge was the result of a last ditch effort by the Germans to turn the tide of battle back in their favor in the West.

The German plan was to drive through the Ardennes and capture the port of Antwerp vital to the Allied war effort. If successful the American and British forces would have to bring supplies all the way from Normandy crippling the Allied war effort.

The American forces defending the area were unprepared for the massive assault of the German forces. The Germans made rapid advances and threatened to achieve a decisive breakthrough.

The stubborn defense of Bastogne by the 101 airborne, the rapid response of Patton's armored units, and the break in the weather allowing the impact of American air superiority combined to stop the German offensive.

The Battle left a large bulge in the Allied lines, but did not break the Allied forces as hoped. The German offensive cost the Germans armored units and elite infantry. The Germans would not be able to launch another offensive after the Battle of the Bulge.

1

Answer:

There are a number of possible answers.

Explanation:

If you mean what is the best example of a primary source document in the study of history, then you are looking for a source which is impartial and objective. This allows the researcher to use it as a valid and reliable source of information from which to draw conclusions.

This could be observations and diaries from a given time in history e.g. Pepys's diaries. They give a personal and detailed insight from someone who lived at that time. However they could be riddled with bias reflected in the observations of the individual thus undermining their validity and reliability.

A second primary source could be a general history from the period, e.g. Tacitus' account of his father-in-law Agricola's campaigns in Britain. Again this gives us an insight from that time but, also again, the account may be highly biased. In this case someone writing about his father-in-law at that time is unlikely to be critical.

A third source is imagery. Early examples would be photographic records of the Crimean War and the American Civil War. In the former such imagery along with the reports by Russell reporting for The Times were instrumental in changing public opinion about the nature of warfare.

A more recent example would be television coverage of the Vietnam War, the first and last extensive TV coverage of conflict. In both instances what is being shown in terms of imagery is metaphorically and literally a snapshot. As with all sources both primary and secondary they have to be treated with caution.

1

Answer:

# y = sqrt(pi)/2 e^(x^2) erf(x) + Ae^(x^2) #

Where #erf(x)# is the Error Function :

# erf(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt #

Explanation:

# dy/dx = 1 + 2xy #
# :. dy/dx - 2xy = 1 # ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

# I = e^(int P(x) dx)#
# \ \ = e^(int \ -2x \ dx)#
# \ \ = e^(-x^2) #

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

# dy/dx - 2xy = 1 #
# :. e^(-x^2)dy/dx - 2xye^(-x^2) = 1*e^(-x^2) #
# :. d/dx(ye^(-x^2)) = e^(-x^2) #

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

# ye^(-x^2) = int \ e^(-x^2) \ dx#

The RHS integral does not have an elementary form, but we can use the definition of the Error Function :

# erf(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt #

Which gives us:

# ye^(-x^2) = sqrt(pi)/2erf(x) + A#
# y = sqrt(pi)/2 e^(x^2) erf(x) + Ae^(x^2) #

1

Answer:

#20#

Explanation:

enter image source here

Given #AB=10, BC=14 and AC=16#,

Let #D,E and F# be the midpoint of#AB,BC and AC#, respectively.

In a triangle, the segment joining the midpoints of any two sides will be parallel to the third side and half its length.

#=> DE# is parallel to #AC, and DE=1/2AC=8#
Similarly, #DF# is parallel to #BC, and DF=1/2BC=7#
Similarly, #EF# is parallel to #AB, and EF=1/2AB=5#

Hence, perimeter of #DeltaDEF=8+7+5=20#

side note : #DE, EF and FD# divide #DeltaABC# into 4 congruent triangles, namely, #DeltaDBE, DeltaADF,DeltaFEC and DeltaEFD#

These 4 congruent triangles are similar to #DeltaABC#

1

Answer:

#-16#

Explanation:

When evaluating an expression with #color(blue)"mixed operations"# there is a particular order that must be followed.

Following the order as set out in the acronym PEMDAS

[Parenthesis(brackets) , Exponents(powers), Multiplication, Division, Addition, Subtraction ]

#rArr8+6xx(4-20)/2^2#

#=8+6xx(-16)/4larrcolor(red)"brackets/powers"#

When multiplication/division are in the calculation, as here, we evaluate from left to right.

#rArr8+(6xx(-16))/4#

#=8+(-96)/4larrcolor(red)" multiplication"#

#=8+(-24)larrcolor(red)" division"#

#=8-24=-16#

2

Answer:

#P = P(t -2tsqrt(1/(4t^2+1)), t^2 +sqrt(1/(4t^2+1))) #

So the parametric equations of P are:

# P_x(t) = t -2tsqrt(1/(4t^2+1)) #
# p_y(t) = t^2 +sqrt(1/(4t^2+1)) #

Explanation:

enter image source here

Our parabola #y=x^2# is parametrised by #x=t# and #y=t^2#, and #A(t,t^2)# is a general point on the parabola, and we give the point #P# the coordinates (to be determined) #P(alpha, beta)#

Differentiating wrt #x# we have:

#dy/dx=2x = 2t #

So the gradient of the tangent at #A# is given by #dy/dx=t#, We are told that #AP# is perpendicular to to parabola, in other words, it is perpendicular to the tangent at A. Hence the gradient of AP is #-1/(2t)# as the product of their gradients are -1.

We can now form the equation of the normal #AP# using the point-slope form of the straight line, #y-y_1=m(x-x_1)#:

# y-t^2=-1/(2t)(x-t) #
# :. 2ty-2t^3=-(x-t) #
# :. 2ty-2t^3=+t-x #

#P(alpha, beta)# lies on this line, giving:

# :. 2tbeta-2t^3 = t -alpha#

And we are also told that #AP# has length 1, so by Pythagoras:

# (t-alpha)^2+(t^2-beta)^2=1^2 #

We can combine these equations to eliminate #t -alpha#:

# (2tbeta-2t^3)^2+(t^2-beta)^2=1 #
# :. 4t^2beta^2-8t^4beta+4t^6 + t^4-2t^2beta+beta^2 = 1 #
# :. (4t^2+1)beta^2-2t^2(4t^2+1)beta +t^4(4t^2+1)=1 #
# :. beta^2-2t^2beta +t^4=1/(4t^2+1) #
# :. beta^2-2t^2beta +t^4-1/(4t^2+1) =0#

We can solve this quadratic in #beta# by completing the square to get:

# (beta-t^2)^2-t^4 +t^4-1/(4t^2+1) =0#
# :. (beta-t^2)^2 = 1/(4t^2+1)#
# :. beta-t^2 = +-sqrt(1/(4t^2+1))#
# :. beta = t^2 +-sqrt(1/(4t^2+1))#

We have found two solutions because there is one point #AP# distance #1# unit extending inwards,and one point extending outwards. Intuitively we choose #beta=t^2 +sqrt(1/(4t^2+1))#.

Now from before we have:

#2tbeta-2t^3 = t -alpha# :
# :. alpha = t -2tbeta+2t^3 #

Substituting our valu of #beta#, from above, gives us:

# alpha = t -2t(t^2 +sqrt(1/(4t^2+1)))+2t^3#
# alpha = t -2t^3 -2tsqrt(1/(4t^2+1))+2t^3#
# alpha = t -2tsqrt(1/(4t^2+1))#

And so the general coordinates of #P(alpha,beta)# in term of #t# are:

#P = P(t -2tsqrt(1/(4t^2+1)), t^2 +sqrt(1/(4t^2+1))) #

So the parametric equations of P are:

# P_x(t) = t -2tsqrt(1/(4t^2+1)) #
# p_y(t) = t^2 +sqrt(1/(4t^2+1)) #

The loci traced out by #P# as #A# moves along the parabola is shown in the following graph in green:

enter image source here

2

Answer:

#f(x) = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))# with radius of convergence #R=1#.

Explanation:

We have:

#f(x) = int_0^x tln(1-t)dt#

Focus on the function:

#ln(1-x)#

We know that:

# ln (1-x) = -int_0^x (dt)/(1-t)#

where the integrand function is the sum of the geometric series:

#sum_(n=0)^oo t^n = 1/(1-t)#

then we can integrate term by term, and we have:

#ln(1-x) = -int_0^x (sum_(n=0)^oo t^n)dt = -sum_(n=0)^oo int_0^x t^ndt =- sum_(n=0)^oo x^(n+1)/(n+1)#

and mutiplying by x term by term:

#xln(1-x) = - sum_(n=0)^oo x^(n+2)/(n+1)#

We can substitute this expression in the original integral and integrate again term by term:

#f(x) = int_0^x tln(1-t)dt = - int_0^x ( sum_(n=0)^oo t^(n+2)/(n+1))dt = -sum_(n=0)^oo int_0^x t^(n+2)/(n+1)dt = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))#

To determine the radius of convergence we can then use the ratio test:

#abs (a_(n+1)/a_n) = abs (frac (x^(n+4)/((n+4)(n+2))) (x^(n+3)/((n+3)(n+1)))) = abs (x) ((n+3)(n+1))/((n+4)(n+2))#

#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)#

and the series is absolutely convergent for #abs(x) <1# which means the radius of convergence is #R=1#