Indefinite Integration by Parts

Find an antiderivative (indefinite integral or primitive) of a product-type function using integration by parts, presenting the result in closed form.

jee-main 2012 Q84 View
If $\displaystyle\int f(x)\,dx = \psi(x)$, then $\displaystyle\int x^{5}f(x^{3})\,dx$ is equal to
(1) $\frac{1}{3}x^{3}\psi(x^{3}) - 3\displaystyle\int x^{3}\psi(x^{3})\,dx + C$
(2) $\frac{1}{3}\left[x^{3}\psi(x^{3}) - \displaystyle\int x^{2}\psi(x^{3})\,dx\right] + C$
(3) $\frac{1}{3}x^{3}\psi(x^{3}) - \displaystyle\int x^{2}\psi(x^{3})\,dx + C$
(4) $\frac{1}{3}\left[x^{3}\psi(x^{3}) - \displaystyle\int x^{3}\psi(x^{3})\,dx\right] + C$
jee-main 2023 Q74 View
For $\alpha , \beta , \gamma , \delta \in \mathbb { N }$, if $\int \left( \frac { x^2 e^x + e^{2x} } { x } \log _ { e } x \right) dx = \frac { 1 } { \alpha } \frac { x^{\beta} e^x } { 1 } - \frac { 1 } { \gamma } \frac { e ^ { \delta x } } { x } + C$, where $e = \sum _ { n = 0 } ^ { \infty } \frac { 1 } { n ! }$ and $C$ is constant of integration, then $\alpha + 2\beta + 3\gamma - 4\delta$ is equal to
(1) 1
(2) 4
(3) $-4$
(4) $-8$