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

Question

Ezra Adams · Accepted Answer

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

Luke Alvarez · Answer

The answer is #=4.43#

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#

William Abdalla · Answer

undefined To solve the integral $ \int_{4}^{9} \frac{x + 1}{x + 2\sqrt{x} - 3} \, dx $, we can start by performing a substitution to simplify the integral. Let $ u = \sqrt{x} $, then $ du = \frac{1}{2\sqrt{x}} \, dx $. Rearranging, we have $ dx = 2u \, du $.

Substituting $ u = \sqrt{x} $ and $ dx = 2u \, du $, we get:

$$ \int \frac{x + 1}{x + 2\sqrt{x} - 3} \, dx = \int \frac{(u^2 + 1) \cdot 2u}{(u^2 + 2u - 3) \cdot 2u} \, du $$
$$ = \int \frac{2u(u^2 + 1)}{2u(u + 3)(u - 1)} \, du $$
$$ = \int \frac{u^2 + 1}{(u + 3)(u - 1)} \, du $$

Now we use partial fraction decomposition to split the integrand:

$$ \frac{u^2 + 1}{(u + 3)(u - 1)} = \frac{A}{u + 3} + \frac{B}{u - 1} $$

Multiplying both sides by $ (u + 3)(u - 1) $, we get:

$$ u^2 + 1 = A(u - 1) + B(u + 3) $$

We solve for $ A $ and $ B $:

$$ u^2 + 1 = Au - A + Bu + 3B $$

$$ u^2 + 1 = (A + B)u + (-A + 3B) $$

Comparing coefficients, we have:

$$ A + B = 0 $$
$$ -A + 3B = 1 $$

Solving these equations, we find $ A = -\frac{1}{4} $ and $ B = \frac{1}{4} $.

Now, we can rewrite the integral as:

$$ \int \left( \frac{-\frac{1}{4}}{u + 3} + \frac{\frac{1}{4}}{u - 1} \right) \, du $$

$$ = -\frac{1}{4} \ln|u + 3| + \frac{1}{4} \ln|u - 1| + C $$

Finally, substitute back $ u = \sqrt{x} $ and $ dx = 2u \, du $:

$$ = -\frac{1}{4} \ln|\sqrt{x} + 3| + \frac{1}{4} \ln|\sqrt{x} - 1| + C $$

Now, evaluate this expression from $ x = 4 $ to $ x = 9 $ and subtract the lower limit result from the upper limit result to get the final answer.