Question

( t2-1) dt integrate it ??​

Answer 1

Answer:

${t}^{3 } \div 3 + t$

Answer 2

$\bf\large\underline\blue{Question:-}$

• Integrate $( {t}^{2} - 1) \: dt$

$\bf\large\underline\blue{Solution:-}$

$\int ( {t}^{2} - 1) \: dt$

Using the property of integrals, i.e.,

$\int f(x) \pm g(x) \: dx = \int f(x) \: dx \pm \int g(x) \: dx$

We get,

$\int {t}^{2} \: dt - \int - 1 \: dt$

Now,

We know that of $t^{n}$ is$\frac{ {t}^{1 + n} }{1 + n}$

Therefore,

$\int {t}^{2} \: dt - \int - 1 \: dt$

$= \frac{ {t}^{3} }{3} - \int - 1 \: dt$

$= \frac{ {t}^{3} }{3} - t$

Now, add the constant of integration, we get,

$= \frac{ {t}^{3} }{3} - t + C, C\in\R$

$\bf\large\underline\blue{Answer:-}$

• $\frac{ {t}^{3} }{3} - t + C, C\in\R$