#### notes

**Integral of the type:**

`int e^x [ f(x) + f'(x)] dx`

we have I = `int e^x[f(x) + f'(x)] dx `

= `int e^x f(x) dx + int e^x f'(x) dx`

= `I_1 + int e^x f'(x) dx , "where" I_1 = int e^x f(x) dx` ...(1)

Taking f(x) and `e^x` as the first function and second function, respectively, in `I_1` and integrating it by parts, we have `I_1`

= `f(x) e^x - int f'(x) e^x dx +C`

Substituting `I_1` in (1), we get

I = `e^x f(x) - int f'(x) e^x dx + int e^x f'(x) dx +C = e^x f(x) + C`

Thus , `int e^x [f(x) + f'(x)] dx = e^x f(x) + C`

**Integrals of some more types :**

Some special types of standard integrals based on the technique of integration by parts :

i) `int sqrt (x^2 - a^2) dx`

`I = int sqrt (x^2 -a^2) dx = x/2 sqrt (x^2 - a^2) - a^2/2 log |x + sqrt (x^2 -a^2)| + C`

ii) `int sqrt (x^2 + a^2) dx`

= `1/2 x sqrt (x^2 + a^2) + a^2/2 log |x + sqrt (x^2 +a^2)| +C`

iii) `int sqrt (a^2 - x^2)dx = 1/2 x sqrt (a^2-x^2) + a^2/2 sin^(-1) x/a+C`

Video link : https://youtu.be/vOQGYpfnC2U