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))
