How do I find the integral
#int(x^3+4)/(x^2+4)dx# ?

Question

Aurora Alvarado · Accepted Answer

#I=1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+c#, where #c# is a constant

Explanation,

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

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

#I=int(x^3)/(x^2+4)dx+int4/(x^2+4)dx #

#I=I_1+I_2# .......#(i)#

Now considering only first integral, which is #I_1=int(x^3)/(x^2+4)dx#

Using Integration by Substitution,

let's #x^2=t# , then #2xdx=dt# yields, first integral

#=intt/2*1/(t+4)*dt#

#=1/2intt/(t+4)*dt#, this can be written as

#=1/2int(t+4-4)/(t+4)*dt#

#=1/2int(1-4/(t+4))dt#

#=1/2intdt-2int1/(t+4)dt#

#=1/2t-2ln(t+4)+c_1#, where #c_1# is a constant

replacing #t#, we get,

#I_1=1/2x^2-2ln(x^2+4)+c_1#, where #c_1# is a constant

considering second integral

#I_2=int4/(x^2+4)dx#

using Trigonometric Substitution to solve this problem,

#I_2=4*1/2tan^-1(x/2)+c_2#

#I_2=2tan^-1(x/2)+c_2#

Finally, plugging in both #I_1# and #I_2# in #(i)#

#I=1/2x^2-2ln(x^2+4)+c_1+2tan^-1(x/2)+c_2#

#I=1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+c#, where #c=c_1+c_2# is again a constant

Christopher Agresta · Answer

undefined To find the integral of $\int \frac{x^3 + 4}{x^2 + 4} \, dx$, you can perform polynomial long division or use the method of partial fractions. After simplifying, the integral can be expressed as the sum of two integrals. The first integral is a straightforward polynomial integral, and the second integral can be solved using a substitution method. Once both integrals are computed, you can sum them up to find the final result.

How do I find the integral #int(x^3+4)/(x^2+4)dx# ?