Indefinite Integral: Finding the indefinite integral of a function involves using integration techniques.
Integration by Parts: A technique based on the formula ( \int u , dv = uv - \int v , du ).
Example 1: ( \int x e^x , dx )
Set ( u = x ), ( du = dx )
Set ( dv = e^x dx ), ( v = e^x )
Apply formula: ( x e^x - \int e^x dx = x e^x - e^x + C )
Example 2: ( \int x \sin x , dx )
Set ( u = x ), ( du = dx )
Set ( dv = \sin x dx ), ( v = -\cos x )
Apply formula: ( -x \cos x + \int \cos x dx = -x \cos x + \sin x + C )
Example 3: ( \int x^2 \ln x , dx )
Choose wisely: ( u = \ln x ), ( du = \frac{1}{x} dx )
( dv = x^2 dx ), ( v = \frac{x^3}{3} )
Simplify: ( \frac{x^3}{3} \ln x - \frac{1}{9} x^3 + C )
Example 4: ( \int \ln x , dx )
Set ( u = \ln x ), ( du = \frac{1}{x} dx )
Set ( dv = dx ), ( v = x )
Result: ( x \ln x - x + C )
Example 5: ( \int x^2 e^x , dx )
Requires multiple uses of integration by parts.
Final result: ( e^x(x^2 - 2x + 2) + C )
Example 6: ( \int \ln x^7 , dx )
Simplified using log properties to ( 7 \int \ln x , dx ).
Result: ( 7(x \ln x - x) + C )
Example 7: ( \int e^{3x} \cos 4x , dx )
Use integration by parts twice.
Final result: ( \frac{3}{25} e^{3x} (\cos 4x + \frac{4}{3} \sin 4x) + C )
Example 8: ( \int \ln x^2/x , dx )
Use integration by parts and properties of logs.
Result: ( \frac{1}{3} (\ln x)^3 + C )
Conclusion
Integration by parts is a powerful tool for solving various integral problems, especially those involving products of algebraic and transcendental functions.