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Determine the integral of ((4)/(21+8x-4x^2))dx by completingthe square, using substitution, table of integrals?
Q. Determine the integral of ((4)/(21+8x-4x^2))dx by completingthe square, using substitution, table of integrals?
Asked by J - Fri Oct 9 17:46:15 2009 - - 1 Answers - 0 Comments

A. Note that by completing the square: 21 + 8x - 4x^2 = 21 - 4(x^2 - 2x) = 21 - 4(x^2 - 2x + 1) + 4 = 25 - 4(x - 1)^2. Let u = x - 1, du = dx. Thus, we get integral 4 du/(25 - 4u^2). Now, use partial fractions: 4/(25 - 4u^2) = A/(5 + 2u) + B/(5 - 2u) ==> 4 = A(5 - 2u) + B(5 + 2u). u = 5/2 ==> 4 = 10B ==> B = 2/5. u = -5/2 ==> 4 = 10A ==> A = 2/5. Thus, we may rewrite the integral as integral (1/5) * [2/(5 + 2u) + 2/(5 - 2u)] du = (1/5) * (ln|5 + 2u| - ln|5 - 2u|) + C = (1/5) * ln|(5 + 2u) / (5 - 2u)| + C = (1/5) * ln|(5 + 2(x - 1)) / (5 - 2(x - 1))| + C. I hope this helps!
Answered by kb - Fri Oct 9 19:27:06 2009

Determine the integral of ((radical(2+(lnx)^2))/(x) )dx use substitution then table of integrals?
Q. Determine the integral of ((radical(2+(lnx)^2))/(x) )dx use substitution then table of integrals?
Asked by J - Fri Oct 9 17:42:31 2009 - - 1 Answers - 0 Comments

A. substitute u = ln(x) <=> x = e^(u) => dx = e^(u) du (2 + ln(x) )/x dx = ( (2 + u )/e^(u)) udu = (2 + u ) du Substitute u = 2 tan( ) => du = 2 (tan ( ) + 1) = 2 sec ( ) d (2 + u ) du = (2 + 2 tan ( )) 2 sec ( ) d = 2 (1 + tan ( )) sec ( ) d = 2 (sec ( )) sec ( ) d = 2 sec ( ) d From integral table [1]: sec ( ) d = sec ( ) sin( )/(n-1) + [(n-2)/(n-1)] sec ( ) d = sec ( ) tan( )/(n-1) + [(n-2)/(n-1)] sec ( ) d and sec ( ) d = ln|sec( ) + tan( )| 2 sec ( ) d = 2 ( sec( ) tan( )/(2) + [(1/(2)] sec( ) d ) = sec( ) tan( ) + ln|sec( ) + tan( )| + c = (1 + tan ( )) tan( ) + ln| (1 + tan ( )) + tan( )| + c substitute back tan( ) = u/ 2 (1 + tan ( )) tan( ) + ln| (1 + tan ( )) + tan( )| + c = (1 + [cont.]
Answered by schmiso - Tue Oct 13 07:47:20 2009

What's the indefinite integral of ln(x)?
Q. I'd like to know what ln(x) is... Scientific workplace says ln(x)dx= x(ln x-1) and I can't find it in my "immediate integrals" table. If I integrate by parts, ln(x)dx = x*ln(x)-1 is this the result I should trust ? Because if I differentiate x*ln(x)-1 I end up with ln(x)+1, which is not ln(x)... So what do I have to differentiate so I can have ln(x) ? What's the indefinite integral of ln(x) ?
Asked by Joao - Fri May 22 11:31:45 2009 - - 3 Answers - 0 Comments

A. ln(x) = 1. ln x = ln x 1 - 1 * d/dx(lnx) = x lnx - x*1/x = x lnx - x +C = x(ln x -1) +C (Ans)
Answered by unknown - Fri May 22 11:40:34 2009