Using the definition of integration for solving integrals, it is quite a long process to solve complex integrals. In this way, mathematicians solve complex integrals and make some integration rules which makes integration easy and interesting for us. So, for evaluating these types of complex integrals we will use the following basic integration rules:
Integral of Function Multiplied by Constant:
The integral of function which is multiplied with a simple constant number is given as follow:
$$ \int k f(x) dx = k \int f(x) dx $$
Proof
The Mathematical proof of the above integral of function is given below:
Suppose that f(x) ia function with the antiderivative of F(x). we can say that
f(x) = F'(x)
Now by using the basic rule of derivative, we can write
$$ (kF(x))' = k F'(x) = k f(x) $$
So kF(x) is an anti derivative of the function kf(x). In simple words
$$ \int k f(x) dx = k F(x) + c = k \int f(x) dx $$
Example
Let's see how to solve integrals having a constant number multiplied with the function under the integral. Consider we have a function
$$ \int 3x^4 dx $$
Solution:
By using our above integral rule of function multiplied by constant, the integral will be:
$$ \int 3x^4 dx = 3 \int x^4 dx$$ $$ \int 3x^4 dx = 3 \left( \frac{x^{4+1}}{4+1} + c_1 \right) $$ $$ \int 3x^4 dx = 3 \left( \frac{x^{5}}{5} + c_1 \right) $$ $$ \int 3x^4 dx = \frac{3}{5} x^5 + (3)c_1 $$ $$ \int 3x^4 dx = \frac{3}{5} x^5 + c $$
Since c1 and 3 both are constant, then they also created a new constant after their multiplication.
Power Rule of Integration:
The power rule of integration just opposite to power rule used in derivatives. Mathematically,
$$ \int x^n dx = \frac{x^{n+1}}{n+1} + c $$
Proof
Since the derivatove of expression xn+1+k is (n+1)xn. Now according to the rule of inverse in differentiation and integration, the andti-derivatove of function (n+1)xn is xn+1+k. Mathematically:
$$ \int (n+1)x^n dx = x^{n+1} + k $$
Now again we use rule of integration for a function multiplied by constant.
$$ \implies (n+1) \int x^n dx = x^{n+1} + k $$
After simplification
$$ \implies \int x^n dx = \frac{x^{n+1} + k}{(n+1)} $$ $$ \implies \int x^n dx = \frac{x^{n+1}}{n+1} + \frac{k}{(n+1)} $$
Thus the rule for solving integral having integrand with some power is proved. For more general, we will replace a constant term k/n+1 by c. Because here k and n are both constants which will produce a new constant. Now,
$$ \implies \int x^n dx = \frac{x^{n+1}}{n+1} + c $$
Example
Evaluate the integral by using power rule of integration:
$$ \int x^5 dx $$
Solution:
$$ \int x^5 dx = x^{5+1}{5+1} + c $$ $$ \int x^5 dx = x^{6}{6} + c $$
Sum and Difference Rule:
If we have two fuction f(x) and g(x) which are adding or subtracting with each other under the integral. Then it may be treated as:
$$ \int \left[ f(x) \pm g(x) \right] dx = \int f(x) dx \pm \int g(x) dx $$
Proof
Suppose that there is a function F(x) which is the anti-derivative of the function f(x) and simlarly we have a function G(x) which is the anti-derivative of the function g(x). So that,
F'(x) = f(x) & G'(x) = g(x)
The basic sum and difference rule of differentiation is given as :
$$ (F(x) \pm G(x)) = F'(x) \pm G'(x) = f(x) \pm g(x) $$
So, F(x)+G(x) is the antiderivative of the function f(x)+g(x) and similarly F(x)-G(x) is the antiderivative of the function f(x)-g(x). Simply,
$$ \int f(x) \pm g(x) dx = F(x) \pm G(x) + c $$ $$ \int f(x) \pm g(x) dx = \int f(x) dx \pm \int g(x) dx $$
Example
Evluate the following integral:
$$ \int \left[ x + e^x \right] dx $$
Solution:
According to the above rule of integration:
$$ \int \left[ x + e^x \right] dx = \int x dx + \int e^x dx $$
Now we will simplify both integrals seprately on the right hand side
$$ \int \left[ x + e^x \right] dx = \frac{x^{1+1}}{1+1} + \frac{e^x}{\frac{d}{dx}(x)} + c $$ $$ \int \left[ x + e^x \right] dx = \frac{x^{2}}{2} + e^x + c $$
Similarly, Let's try for subtractio rule of integration
Evaluate the following integral:
$$ \int \left[ 2 - \frac{1}{x} \right] dx $$
Solution:
According to the above rule of integration:
$$ \int \left[ 2 - \frac{1}{x} \right] dx = \int 2 dx - \int \frac{1}{x} dx $$ $$ \int \left[ 2 - \frac{1}{x} \right] dx = 2 \int 1 dx - \int \frac{1}{x} dx $$ $$ \int \left[ 2 - \frac{1}{x} \right] dx = 2(x) - ln(x) + c $$
Since, integral of 1/x is ln(x)
Integration by Parts:
From the list of all integration rules, Integration by parts is on eof the most important and basic rule for solving integrals. In integration by parts there are two different functions i.e "u" and "v" are multiplied by each other under the integral. So, the integration by parts will be:
$$ \int uv \; dx = u \int v dx - \int u' (\int v dx) dx $$
The major thing in integration by parts is choosing a right “u” and “v”.
Integration by Parts Example
Evaluate the integral :
$$ \int x cos(x) dx $$
Solution:
The first step in solving of two function multiplied with each other is choosing right "u" and "v".
Here,
Consider u = x & v = cos(x)
Now, in second step differentiate "u" and integrate "v".
u = x
⇒ u' = (x)' = 1
Now,
v = x
∫v = ∫cos(x) dx = sin(x)
In the last step, we have to simply putting all these values in the formula of integration by parts example:
$$ \int uv \; dx = u \int v dx - \int u' (\int v dx) dx $$ $$ x sin(x) - \int 1(sin(x)) dx $$ $$ x sin(x) - \int (sin(x)) dx $$ $$ x sin(x) + cos(x) + c $$
Now thats end and we have successfully solved integration by parts example.
So, these are some basic integration rules that provides you a smooth way for evaluating easy and complex integrals.