To solve the fraction integral function you can use the chain rule which gives you substitution and the product rule which gives you integration by parts.
Using these both rules you can solve or show your final solution according to this result-
From this formula, you can get the idea of what you need to further integrate the divisible function.
Or you can use the integration by parts rule to solve division integral functions by taking one function as u and the other as v according to the ILATE rule.
I = Inverse trigonometric function
L = Logarithmic function
A = Algebraic function
T = Trigonometric function
E = Exponential Function
Integration by parts u/v form-
\int _ { } ^ { } udv = u.v – \int _ { } ^ { } v.du
Note: There is no direct rule which solve the fraction or division integral problem as product rule does.
We use different-different methods to cope with this type of problem like Integration Partial fraction for repeated or non repeated linear factor, Integration Substitution Method, Integration by parts, etc.
Indefinite Integration of a quotient polynomial using substitution examples
Example 1: \int _ { } ^ { } \frac { z + 5 } { z ^ { 2 } + 10z } dz
Solution-
⇒ u = z² + 10z
⇒ du = (2z + 5)dz
Divide by 2, L.H.S and R.H.S
⇒ \frac { du } { 2 } = \frac { 2 ( z + 5 ) dz } { 2 }
⇒ \frac { du } { 2 } = ( z + 5 ) dz
Put all vaules in problem
= \frac { 1 } { 2 } \int _ { } ^ { } \frac { du } { u }
= \frac { 1 } { 2 } ln|u| + C
= \frac { 1 } { 2 } ln|z ^ { 2 } + 10z | + C Answer
Example 2: \int _ { } ^ { } \frac { 3z + 7 } { z ^ { 2 } + 4 } dz
Solution-
Separate the function in such way we get same denominator after it
= 3 \int _ { } ^ { } \frac { z } { z ^ { 2 } + 4 } dz + 7 \int _ { } ^ { } \frac { 1 } { z ^ { 2 } + 4 } dz
Now substitution:
⇒ u = z² + 4
⇒ du = 2z dz
Take this function \int _ { } ^ { } \frac { z } { z ^ { 2 } + 4 } dz
and substitute the above value in given function
= \int _ { } ^ { } \frac { \frac { 1 } { 2 } du } { u } = 1/2 ln|u| + C
= 1/2 ln (z² + 4) + C Answer
Also check –