# find the integral: 1.   [x - (5/x) -e ^-2x]dx 2.  (-sinx / cosx)dx 3.   (2t - 9)^4 dt 4.   (6x - 10) / (x^2 - 2x -3) dxfor equation 4 use partial fractions

Asked on by donovanp

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

1. [x - (5/x) -e ^-2x]dx

intg[x - (5/x) - e^-2x] dx = intg xdx-5*intg(1/x)dx - intg(e^-2x)dx

= (x^2/2) - 5lnx - (e^-2x/-2) + C

2.  (-sinx / cosx)dx

intg ( -sinx/cosx)dx = intg (-tanx) dx

= - intg(tanx)dx

= -(-ln l cosx l ) + C

= ln l cosx l + C

3.   (2t - 9)^4 dt

Let u= 2t- 9   ==> du = 2 dt

intg (2t -9)^4 dt = intg u^4 du

=( u^5/5 ) du

= (2t-9)^5/5*2

= (1/10)*(2t-9)^5 + C

4.   (6x - 10) / (x^2 - 2x -3) dx

intg (6x-10)/(x-3)(x+1) dx = intg (6x-10)/(x-3)(x+1) dx

==> A/(x-3) + b/(x+1) = (6x-10)/(x-3)(x+1)

==> A(x+1) + B(x-3) = 6x - 10

==> Ax + A + Bx - 3B = 6x -10

==> (A+B)x + (A-3B) = 6x - 10

==> A+B = 6

==> A-3B = -10

==> A = 3B -10

==? 3B -10 + B = 6

==> 4B = 16

=> B = 4

==> A = 2

==> intg (2/(x-3) dx + intg (4/(x+1) dx

= 2ln (x-3) + 4 ln (x+1)  + C

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Because the 4th integral is more elaborated, I''ll start by solving it:

We notice that if we note the denominator

x^2 - 2x -3 = t => (2x-2)dx = dt

The numerator of the ratio is 6x - 10.

The numerator could be written as:

6x - 10 = 2x-2 + 4x-8

The ratio will be written:

(6x - 10) / (x^2 - 2x -3)=(2x-2)/(x^2 - 2x -3)+(4x-8)/(x^2 - 2x -3)

Int(6x - 10) dx/ (x^2 - 2x -3)=Int(2x-2)dx/(x^2 - 2x -3)+Int(4x-8)dx/(x^2 - 2x -3)

We'll solve Int(2x-2)dx/(x^2 - 2x -3)

Int(2x-2)dx/(x^2 - 2x -3) = Int dt/t = ln t + C

Int(2x-2)dx/(x^2 - 2x -3) = ln (x^2 - 2x -3) + C

We'll solve Int(4x-8)dx/(x^2 - 2x -3).

Int(4x-8)dx/(x^2 - 2x -3) = ln (x^2 - 2x -3) -4 Int dx/(x^2 - 2x -3)

We'll solve Intdx/(x^2 - 2x -3)

We'll write the denominator as a product of linear factors.

The roots of the equation x^2 - 2x -3 = 0 are:

x1+x2 = -2

x1*x2 = -3

x1 = -3 and x2 = 1

The equation will be written:

x^2 - 2x -3 = (x-x1)(x-x2) = (x+3)(x-1)

Now, we'll write the ratio:

1/ (x^2 - 2x -3) = 1/(x+3)(x-1)

as a sum of simple irreducible ratios.

1/(x+3)(x-1) = a/(x+3) + b/(x-1)

1 = a(x-1) + b(x+3)

We'll remove the brackets:

1 = ax - a + bx + 3b

We'll combine like terms:

1 = x(a+b) + 3b-a

a+b = 0 => -a = b

3b-a = 1 => 3b+b = 1 => 4b = 1 => b = 1/4 => a = -1/4

1/(x+3)(x-1) = -1/4(x+3) + 1/4(x-1)

Int dx/(x+3)(x-1) = (-1/4)*Int dx/(x+3) + (1/4)*Intdx/(x-1)

Int dx/(x+3) = ln (x+3) + C

Intdx/(x-1) = ln (x-1) + C

Int dx/(x+3)(x-1) = (1/4)[ln (x-1) -  ln (x+3)] + C

Int dx/(x+3)(x-1) = (1/4)*ln [(x-1)/(x+3)]  + C

Int dx/(x+3)(x-1) = ln [(x-1)/(x+3)] + C

The result of integral is:

Int (6x - 10) dx/ (x^2 - 2x -3) = 3ln (x^2 - 2x -3) - ln [(x-1)/(x+3)] + C

Int (6x - 10) dx/ (x^2 - 2x -3) = ln [(x+3)*(x^2 - 2x -3)^3/(x-1)] + C

neela | High School Teacher | (Level 3) Valedictorian

Posted on

1)

Using the standard integratios we do the integration.

Int {x-5/x-e^-2x}dx = Intxdx -5*dx/x -int e^(2x) dx

= x^2/2 -5lnx- e^(-2x)/(-2), as inx^ndx = x^(n+1) /2; Int ln dx/x = lnx and  Int e^ (kx) dx = e^(kx) /k

= [x^2 +e^(-2x)}/2 - 5lnx + constant

2)

Int(-sinx/cosx )dx = Int d(cosx)/(cosx ) +constant= ln(cosx)+ constant

3)

Int(2t-9)^4 dt = Int y^4 *dy /2, where y = 2t-7. So dt = dy/2.

=[ y^5/5](1/2) +Constant

=0.1(2t-9)^5 + Constant.

4)

Int (6x-10)/(x^2-2x-3) dx .

We write the integrand into sum of two partial fractions as below:

(6x-10)/(x-3)(x+1) =  [6(3) -10)]/4(x-3) + [6(-1)-10]/(-4)(x+1)

= 2/(x-3) + 4/(x+1).

So Int [(6x-10)/(x^2-2x-3)]dx = Int dx/(x-3) +4dx/(x+1)

= int 2*d(x-3)/(x-3) + int 4*d(x+1)/(x+1)

= 2ln(x-3) +4ln(x+1)

= 2{ln(x-3)+2ln(x+1)}

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