∫ (4+x)² / (x²+4) please help me solve this? :(

Question

∫ ((4+x)² / (x²+4))

Isaiah Allen · Accepted Answer

The answer is #=x+4ln(x^2+4)+6arctan(x/2)+C#

We need

#int(dx)/(x^2+1)=arctan(x)+C#

Perform a polynomial long division

#(4+x)^2/(x^2+4)=(x^2+8x+16)/(x^2+4)=1+(8x+12)/(x^2+4)#

#=1+(8x)/(x^2+4)+12/(x^2+4)#

Therefore,

#int((4+x)^2dx)/(x^2+4)=int1dx+4int(2xdx)/(x^2+4)+12int(dx)/(x^2+4)#dx

#=x+4ln(x^2+4)+12int(dx)/(4((x/2)^2+1))#

#=x+4ln(x^2+4)+3int(dx)/((x/2)^2+1)#

#=x+4ln(x^2+4)+6arctan(x/2)+C#

Ezra Adams · Answer

undefined To solve the integral of (4 + x)^2 / (x^2 + 4), we can first rewrite the numerator as (4 + x)^2 = (4 + x)(4 + x) and then expand it using the distributive property. This gives us 16 + 8x + x^2.

Now, we can rewrite the integral as the sum of two separate integrals:

∫ (16 + 8x + x^2) / (x^2 + 4) dx

Next, we can divide each term in the numerator by x^2 + 4:

∫ (16 / (x^2 + 4)) + (8x / (x^2 + 4)) + (x^2 / (x^2 + 4)) dx

Now, we have three separate integrals to solve:

∫ (16 / (x^2 + 4)) dx
∫ (8x / (x^2 + 4)) dx
∫ (x^2 / (x^2 + 4)) dx

The first integral can be solved using the arctangent function:

∫ (16 / (x^2 + 4)) dx = 4 * arctan(x/2) + C1

For the second integral, we can use a substitution. Let u = x^2 + 4, then du = 2x dx:

∫ (8x / (x^2 + 4)) dx = 4 ∫ (1 / u) du = 4 ln|u| + C2

Substitute back u = x^2 + 4:

4 ln|x^2 + 4| + C2

For the third integral, we can use a similar substitution. Let u = x^2 + 4, then du = 2x dx:

∫ (x^2 / (x^2 + 4)) dx = 1/2 ∫ (2x / (x^2 + 4)) dx

= 1/2 ln|x^2 + 4| + C3

Therefore, the final solution to the integral is:

4 * arctan(x/2) + 4 ln|x^2 + 4| + (1/2) ln|x^2 + 4| + C

where C is the constant of integration.