Question

evaluate the indefinite integral.
\\(\int te^{9t^{2}}dt\\)

Explanation:

Step1: Use substitution

Let $u = 9t^{2}$, then $du=18t\ dt$, and $t\ dt=\frac{1}{18}du$.

Step2: Rewrite the integral

The integral $\int te^{9t^{2}}dt$ becomes $\int e^{u}\cdot\frac{1}{18}du$.

Step3: Integrate with respect to u

$\int e^{u}\cdot\frac{1}{18}du=\frac{1}{18}\int e^{u}du$. Since $\int e^{u}du = e^{u}+C$, we have $\frac{1}{18}e^{u}+C$.

Step4: Substitute back u

Substitute $u = 9t^{2}$ back, we get $\frac{1}{18}e^{9t^{2}}+C$.

Answer:

$\frac{1}{18}e^{9t^{2}}+C$