Find:

Solution (a) Notice that the numerator equals the derivative of the denominator, except for a constant factor. So, we try substitution. Let $\ u=4x^{3}-1.$ Then $du=12x^{2}dx$ so $5x^{2}dx=\dfrac{5}{12}\,du$. $$\int \dfrac{5x^{2}dx}{4x^{3}-1}=\int \dfrac{\dfrac{5}{12}\,du}{u}=\dfrac{5}{12}\int \dfrac{du}{u}=\dfrac{5}{12}\ln \left\vert u\right\vert +C=\dfrac{5}{12}\ln \left\vert 4x^{3}-1\right\vert +C$$

(b) Here, the numerator equals the derivative of the denominator. So, we use the substitution $u=e^{x}+4$. Then $du=e^{x}dx$. $$\begin{eqnarray*} \int \dfrac{e^{x}}{e^{x}+4}dx=\int \dfrac{1}{e^{x}+4}\cdot e^{x}dx=\int \dfrac{1}{u}\,du=\ln \vert u\vert +C \underset{\underset{{\color{#0066A7}{\hbox{\(u=e^{x}+4>0\)}}}}{\color{#0066A7}{\uparrow}}}{=} \ln (e^{x}+4) +C \\ \end{eqnarray*}$$