Question

determine the following indefinite integral.
int \frac{40t^{8}-7}{t}dt
int \frac{40t^{8}-7}{t}dt=square

Explanation:

Step1: Split the fraction

$\int\frac{40t^{8}-7}{t}dt=\int(40t^{7}-\frac{7}{t})dt$

Step2: Integrate term - by - term

$\int(40t^{7}-\frac{7}{t})dt=\int40t^{7}dt-\int\frac{7}{t}dt$
For $\int40t^{7}dt$, using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\int40t^{7}dt=40\times\frac{t^{7 + 1}}{7+1}=5t^{8}$.
For $\int\frac{7}{t}dt$, since $\int\frac{1}{x}dx=\ln|x|+C$, then $\int\frac{7}{t}dt=7\ln|t|$.

Step3: Add the constant of integration

$5t^{8}-7\ln|t|+C$

Answer:

$5t^{8}-7\ln|t|+C$