Question

(2) Calculate the following integrals: х 2.x3 – 4x + 3 -dx (x + 1)2(x2 +1) Java 2 - 25 2 dx x4V x2 – 25 (3) Explain why, using the techniques we've learned so far, we are able to calculate the integral of any rational function. (A rational function is one of the form p(x) where g(x2) p and q are polynomials.)

Answer 1

__Part 1)__

Here we have to solve the following given integrals:

Given:

$\int \dfrac{x^4-2x^3-4x+3}{(x+1)^2(x^2+1)}dx\\$

We solve the above integral using partial fraction decomposition emthod.

2.1 4.1 +3 -dc 24 +223 +2.02 +2.0 + 1

As here we can see the degree of numerator is not less than degree of denominator.Hence we apply here

polynomial long division method:

$\dfrac{x^4-2x^3-4x+3}{x^4+2x^3+2x^2+2x+1}=1+\dfrac{-4x^3-2x^2-6x+2}{x^4+2x^3+2x^2+2x+1}$

We can write above equation as:

$1+\dfrac{-4x^3-2x^2-6x+2}{x^4+2x^3+2x^2+2x+1}=1+\dfrac{-4x^3-2x^2-6x+2}{(x+1)^2(x^2+1)}-----eqn(1)$

Here writing eqn(1) in partial fraction decomposition method:

$\dfrac{-4x^3-2x^2-6x+2}{(x+1)^2(x^2+1)}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{Cx+D}{x^2+1}----eqn(2)\\ \dfrac{-4x^3-2x^2-6x+2}{(x+1)^2(x^2+1)}=\dfrac{A(x+1)(x^2+1)+B(x^2+1)+(Cx+D)(x+1)^2}{(x+1)^2(x^2+1)}\\ -4x^3-2x^2-6x+2=Ax^3+Ax+Ax^2+A+Bx^2+B+Cx^3+2Cx^2+Cx+Dx^2+2Dx+D$

Now comparing the coefficients of constant terms from both side we get:

A+B+D=2-----------eqn(3)

Now comparing the coefficients of "
T
" from both side we get:

A+C+2D= -6---------eqn(4)

Now comparing the coefficients of "$x^2$" from both side we get:

A+B+2C+D = -2---------eqn(5)

Now comparing the coefficients of "$x^3$" from both side we get:

A+C = -4---------------eqn(6)

Solving eqn(3),eqn(4),eqn(5) and eqn(6) we get:

A=-2

B=5

C=-2

D=-1

Hence the eq(2) will be now written as:

$\dfrac{-4x^3-2x^2-6x+2}{(x+1)^2(x^2+1)}=\dfrac{-2}{x+1}+\dfrac{5}{(x+1)^2}+\dfrac{-2x-1}{x^2+1}$

Again substituting above value in eq(1) and then integrating it:

$\int \dfrac{x^4-2x^3-4x+3}{(x+1)^2(x^2+1)}dx=\int 1\ dx+\int \dfrac{-2}{x+1}\ dx+\int \dfrac{5}{(x+1)^2}dx+\int \dfrac{-2x-1}{x^2+1}dx\\ =\int 1\ dx+\int \dfrac{-2}{x+1}\ dx+\int \dfrac{5}{(x+1)^2}dx+\int \dfrac{-2x}{x^2+1}dx+\int \dfrac{-1}{x^2+1}dx\\$

Formula we will use below has shown here:

$\int \dfrac{1}{x}dx=ln\left | x \right |+c\\ \int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\ \int \dfrac{1}{x^2+1}dx=tan^{-1}(x)+c$

Let:

Therefore integrating:

=$=x-2ln\left | x+1 \right |+5\left ( \dfrac{-1}{x+1} \right )-ln\left | x^2+1 \right |-tan^{-1}(x)+c$

Here c represents an arbitrary constant.

__Part 2)__

Given:

$\int \dfrac{2}{x^4\sqrt {x^2-25}}dx$----------eqn(1)

Here we are using trigonometric substitution method.

Substituting this in eqn(1) we get:

$\int \dfrac{2}{5^4\ sec^4(z)\sqrt {25\ sec^2(z)-25}}\cdot 5\ tan(z)\ sec(z)\ dz\\ =\int \dfrac{2}{5^4\ sec^4(z)\ 5\sqrt {sec^2(z)-1}}\cdot 5\ tan(z)\ sec(z)dz$

We know the formula of trigonometric:

Therefore:

$=\int \dfrac{2\cdot 5\ tan(z)\ sec(z)}{5^4\ sec^4(z)\ 5\ tan(z)}dz\\ =\int \dfrac{2}{625\ sec^3(z)}dz\\ =\dfrac{2}{625}\int cos^3(z)dz\\ =\dfrac{2}{625}\int cos^2(z)\cdot cos(z)dz\\ =\dfrac{2}{625}\int (1-sin^2(z))\ cos(z)\ dz\\$

Let:

sin(z)=u

cos(z) dz=du

Hence:

$=\dfrac{2}{625}\int (1-u^2)du\\ =\dfrac{2}{625}\int du-\dfrac{2}{625}\int u^2\ du\\ =\dfrac{2}{625}\left [ u \right ]-\dfrac{2}{625}\dfrac{u^3}{3}+c\\$

Substituting back the value of u and z then we get the required solution:

$=\dfrac{2}{625}\left [ sin\left ( sec^{-1}\left ( \dfrac{x}{5} \right ) \right ) \right ]-\dfrac{2}{625*3}\left [ sin^3\left ( sec^{-1}\left ( \dfrac{x}{5} \right ) \right ) \right ]+c\\$

Here c is denoting integration constant.

__Part 3)__

Here we are using the method that is more suitable for integrating any function which is in fraction.

We can too calculate it by integrating it directly but it may consume more time or may be we can stuck during solving.

Here partial fraction decomposition method that allow us to break the difficult problem in simpler one.