1. Choose $u=x$, $dv=e^x dx$. 2. Find $du=dx$, $v=e^x$. 3. Apply formula: $xe^x - \int e^x dx$. 4. Evaluate: $xe^x - e^x + C$.

1. Rewrite as $\int 1 \cdot \ln(x) , dx$. 2. Choose $u=\ln(x)$, $dv=dx$. 3. Find $du=\frac{1}{x}dx$, $v=x$. 4. Apply formula: $x\ln(x) - \int x(\frac{1}{x}) dx$. 5. Evaluate: $x\ln(x) - x + C$.

1. Choose $u=x^2$, $dv=\cos(x)dx$. 2. Find $du=2x dx$, $v=\sin(x)$. 3. Apply formula: $x^2\sin(x) - \int 2x\sin(x) dx$. 4. Apply IBP again to $\int 2x\sin(x) dx$. 5. Final result: $x^2 \sin(x) + 2x \cos(x) - 2\sin(x) + C$.

1. Apply IBP twice, keeping $e^x$ as part of 'u' or 'dv' consistently. 2. Isolate the original integral using algebraic manipulation. 3. Solve for the original integral.

$\int u , dv = uv - \int v , du$

$\frac{d}{dx} uv = u cdot v' + v cdot u'$

To simplify integrals involving the product of two functions by transforming them into simpler integrals.

It's the reverse process of the product rule for differentiation.

Proper selection simplifies the integral, making it easier to solve.