# Evaluate the integral: integral {t (square root t) + t (square root t)} / {t^2} dt.

## Question:

Evaluate the integral:

{eq}\displaystyle \int \dfrac {t (\sqrt t) + t (\sqrt t)} {t^2}\ dt {/eq}.

## Integration:

Integration is Anti-differentiation. Differentiation is used to find the slope of a curve whereas we use integration to find the area under the curve.

Indefinite Integral:

Any integral of the form {eq}\int g(x)dx {/eq} is called indefinite integral in which the upper and the lower limit of the integrand is not unknown.

Given

{eq}\displaystyle \int \dfrac {t (\sqrt t) + t (\sqrt t)} {t^2}\ dt {/eq}.

We have to evaluate the integral.

{eq}\begin{align} \displaystyle \int \dfrac {t (\sqrt t) + t (\sqrt t)} {t^2}\ dt &=\int \dfrac{t^{\dfrac{3}{2}}+t^{\dfrac{3}{2}}}{t^2}dt\ & \left [ \because a^{m}.a^{n}=a^{m+n} \right ]\\ \\ &=\int \dfrac{2t^{\dfrac{3}{2}}}{t^2}dt\\ \\ &=2\int \dfrac{t^{\dfrac{3}{2}}}{t^2}dt\\ \\ &=2\int t^{\dfrac{3}{2}-1}dt \ & \left [ \because \frac{a^{m}}{a^{n}}=a^{m-n} \right ]\\ \\ &=2\int t^{\dfrac{-1}{2}dt}\\ \\ &=2\left [ \dfrac{t^{\dfrac{-1}{2}+1}}{\frac{-1}{2}+1} \right ]\ & \left [ \because \int t^ndt=\frac{t^{n+1}}{n+1} \right ]\\ \\ &=2\left [ \frac{t^{\dfrac{1}{2}}}{\dfrac{1}{2}} \right ]+C\\ \\ &=4\left ( t^{\dfrac{1}{2}} \right )+C \end{align} {/eq}

{eq}\color{blue}{\displaystyle \int \dfrac{t (\sqrt t)+t(\sqrt t)}{t^2}\ dt =4\left ( t^{\dfrac{1}{2}} \right )+C} {/eq} 