Expand or match the inner derivative

$\int\sqrt{x}(1-x)\,dx=\int(x^{1/2}-x^{3/2})\,dx=\frac23x^{3/2}-\frac25x^{5/2}+C$, for $x>0$ (with a right-hand endpoint value at $x=0$).

Use $(a+b)^2=a^2+2ab+b^2$: $\int(x^{1/2}+3)^2\,dx=\int(x+6x^{1/2}+9)\,dx=\frac12x^2+6(\frac23)x^{3/2}+9x+C=\frac12x^2+4x^{3/2}+9x+C$, on $x>0$.

Take $f(x)=5x+4$, $f^{\prime}(x)=5$. Then $\int(5x+4)^5\,dx=\frac15\int5(5x+4)^5\,dx=\frac15\frac{(5x+4)^6}{6}+C=\frac{(5x+4)^6}{30}+C$.

Take $f(x)=2x^2-3$, $f^{\prime}(x)=4x$. Then $\int x(2x^2-3)^4\,dx=\frac14\int4x(2x^2-3)^4\,dx=\frac14\frac{(2x^2-3)^5}{5}+C=\frac{(2x^2-3)^5}{20}+C$.

Take $f(x)=x^2+3x$, $f^{\prime}(x)=2x+3$. Thus $\int\frac{2x+3}{(x^2+3x)^3}\,dx=\int(2x+3)(x^2+3x)^{-3}\,dx=\frac{(x^2+3x)^{-2}}{-2}+C=-\frac1{2(x^2+3x)^2}+C$, on intervals excluding $x=-3,0$.