Sometimes it may happen, especially with products of exponential and trigonometric functions, that executing two integrations by parts

$$\int f(x)\,\mathrm{d}x=\cdots=g(x)-\int f(x)\,\mathrm{d}x$$

In this case, executing more integrations by parts will not solve the integral; instead it is ready to be solved from the previous expression as if it was a completely normal equation

$$\int f(x)\,\mathrm{d}x=\dfrac{1}{2}g(x)+k$$

Depending on the integral, positive constants (usually squared) may accompany the initial integral; this is no inconvenience for the integral to be solved

$$\int f(x)\,\mathrm{d}x=\cdots=g(x)-a\int f(x)\,\mathrm{d}x$$ $$\int f(x)\,\mathrm{d}x=\dfrac{1}{1+a}g(x)+k$$

**keeping the same structure for $u$ and $v$**returns the initial integral, with a change of sign, together with a remainder function:$$\int f(x)\,\mathrm{d}x=\cdots=g(x)-\int f(x)\,\mathrm{d}x$$

In this case, executing more integrations by parts will not solve the integral; instead it is ready to be solved from the previous expression as if it was a completely normal equation

$$\int f(x)\,\mathrm{d}x=\dfrac{1}{2}g(x)+k$$

Depending on the integral, positive constants (usually squared) may accompany the initial integral; this is no inconvenience for the integral to be solved

$$\int f(x)\,\mathrm{d}x=\cdots=g(x)-a\int f(x)\,\mathrm{d}x$$ $$\int f(x)\,\mathrm{d}x=\dfrac{1}{1+a}g(x)+k$$