How do you find #\int _ { 4} ^ { 9} \frac { x + 1} { x+ 2\sqrt { x } - 3} d x#?

2 Answers
Narad T.
Mar 27, 2018

The answer is #=4.43#

Explanation:

Calculate the indefinite integral first

Let #u=sqrtx#, #=>#, #du=1/(2sqrtx)dx#

#I=int((x+1)dx)/(x+2sqrtx-3)=int(2u(u^2+1)du)/(u^2+2u-3)#

#(2u(u^2+1))/(u^2+2u-3)=2(u-2)+(2(8u-6))/(u^2+2u-3)#

#=2(u-2)+(4(4u-3))/(u^2+2u-3)#

Perform the decomposition into partial fractions

#((4u-3))/(u^2+2u-3)=(4u-3)/((u-1)(u+3))#

#=A/(u-1)+B/(u+3)#

#=(A(u+3)+B(u-1))/((u-1)(u+3))#

The denominators are the same, compare the numerators

#4u-3=A(u+3)+B(u-1)#

Let #u=1#, #=>#, #1=4A#

let #u=-3#, #=>#, #-15=-4B#

Therefore,

#2(u-2)+(4(4u-3))/(u^2+2u-3)=2(u-2)+1/(u-1)+15/(u+3)#

And finally

#I=2int(u-2)du+int(du)/(u-1)+int(15du)/(u+3)#

#=u^2-4u+ln(u-1)+15ln(u+3)#

#=x-4sqrtx+ln(sqrtx-1)+15ln(sqrtx+3)+C#

And the definite integral is

#int_4^9((x+1)dx)/(x+2sqrtx-3)=#

#=[x-4sqrtx+ln(sqrtx-1)+15ln(sqrtx+3)]_4^9#

#=(9-12+ln2+15ln6)-(4-8+ln1+15ln5)#

#=15ln6-15ln5+ln2+1#

#=4.43#

Cem Sentin
Mar 27, 2018

#1+15Ln6+Ln2-15Ln5#

Explanation:

#int_4^9 (x+1)/(x+2sqrtx-3)*dx#

After using #y=sqrtx#, #x=y^2# and #dx=2y*dy# transforms, this integral became

#int_2^3 (y^2+1)/(y^2+2y-3)*2y*dy#

=#int_2^3 (2y^3+2y)/(y^2+2y-3)*dy#

=#int_2^3 (2y-4)*dy#+#int_2^3 (16y-12)/(y^2+2y-3)*dy#

=#[y^2-4y]_2^3#+#int_2^3 (16y-12)/((y+3)*(y-1))*dy#

=#1+15int_2^3 dy/(y+3)+int_2^3 dy/(y-1)#

=#1+[15Ln(y+3)+Ln(y-1)]_2^3#

=#1+15Ln6+Ln2-15Ln5#

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