First we need to do partial fractions. We know that the numerator will be of a degree lower than the denominator, so we can write an equation like this:
#1/((x+7)(x^2+9))=A/(x+7)+(Bx+C)/(x^2+9)#
To do partial fractions, we multiply both sides by the left hand side denominator:
#1/cancel((x+7)(x^2+9))cancel((x+7)(x^2+9))=(x+7)(x^2+9)(A/cancel(x+7)+(Bx+C)/cancel(x^2+9))#
This gives us:
#1=A(x^2+9)+(Bx+C)(x+7)#
#1=Ax^2+9A+Bx^2+7Bx+Cx+7C#
If we group coefficients of the same degree, we have:
#1=(A+B)x^2+(7B+C)x+(9A+7C)#
We know how much of each degree of #x# is on the left, so we can setup the following system of equations:
#A+B=0#
#7B+C=0#
#9A+7C=1#
Solving it, we get:
#A=1/58#
#B=-1/58#
#C=7/58#
We can now plug this back into our integral to get:
#int\ (1/58)/(x+7)+(-1/58x+7/58)/(x^2+9)\ dx#
To keep track, I will name the left integral Integral 1 and the right one Integral 2
Integral 1
#int\ (1/58)/(x+7)\ dx=1/58int\ 1/(x+7)\ dx#
We can solve it by letting #u=x+7# and #(du)/dx=1#:
#1/58int\ 1/u\ du=1/58ln|x+7|#
Integral 2
#int\ (-1/58x+7/58)/(x^2+9)\ dx=1/58int\ (-x+7)/(x^2+9)\ dx#
We can split this into two:
#=1/58(-int\ x/(x^2+9)+int\ 7/(x^2+9)\ dx)#
I will call the left one Integral 3, and the right one Integral 4.
Integral 3
Let #u=x^2+9# and #(du)/dx=2x# and divide through by #2x#:
#-int\ x/(x^2+9)\ dx=-int\ 1/(2cancelx)cancelx/u\ du=-1/2int\ 1/u\ du=-1/2ln|u|#
Now we resubstitute, and since #x^2+9# is always positive, we can remove the absolute value part:
#=-1/2ln(x^2+9)#
Integral 4
#int\ 7/(x^2+9)\ dx=7int\ 1/(x^2+9)\ dx#
We want to turn this integral into the familiar form:
#int\ 1/(x^2+1)\ dx=tan^-1(x)#
To do this, we want to factor out a #1/9# from the bottom, so I will let #x=sqrt9u=3u# and #dx/(du)=3#. Note that since we took the derivative with respect to #u#, we have to multiply by it instead of dividing:
#7int\ 1/(x^2+9)\ dx=7int\ 3*1/((3u)^2+9)\ du=21int\ 1/(9u^2+9)\ du#
Now we can factor out on the bottom:
#21int\ 1/(9u^2+9)\ du=21/9int\1/(u^2+1)\ du=21/9tan^-1(u)#
Solving for #u# in #x=3u#, we get #u=x/3#. If we resubstitute, we get:
#21/9tan^-1(x/3)#
Completing Integral 2
Now we can plug in Integral 3 and 4 to get:
#1/58int\ (-x+7)/(x^2+9)\ dx=1/58(-1/2ln(x^2+9)+21/9tan^-1(x/3))#
Completing the original integral
We know what Integral 1 and 2 is equal to, so we can solve the original integral:
#int\ (1/58)/(x+7)+(-1/58x+7/58)/(x^2+9)\ dx=#
#=1/58ln|x+7|+1/58(-1/2ln(x^2+9)+21/9tan^-1(x/3))#
Simplifying, we get:
#=1/58(ln|x+8|+7/3tan^-1(x/3)-1/2ln(x^2+9))#
