What is the integral of (3(1+t^{1/2})^2)/(2t^{1/2)} ?

The integral of (3(1+t^{1/2})^2)/(2t^{1/2)} is 3t^{1/2}+3t+t^{3/2}+C

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integral of (3(1+t^{1/2})^2)/(2t^{1/2)}

\int \frac{3 ( 1 + t ^{\frac{1}{2}} ) ^{2}}{2 t ^{\frac{1}{2}}} d t

3 t^{\frac{1}{2}} + 3 t + t^{\frac{3}{2}} + C

Solution steps

\int \frac{3 ( 1 + t ^{\frac{1}{2}} ) ^{2}}{2 t ^{\frac{1}{2}}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{2} \cdot \int \frac{( 1 + t ^{\frac{1}{2}} ) ^{2}}{t ^{\frac{1}{2}}} d t

Expand

\frac{( 1 + t ^{\frac{1}{2}} ) ^{2}}{t ^{\frac{1}{2}}} : \frac{1}{t ^{\frac{1}{2}}} + 2 + t^{\frac{1}{2}}

= \frac{3}{2} \cdot \int \frac{1}{t ^{\frac{1}{2}}} + 2 + t^{\frac{1}{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{3}{2} (\int \frac{1}{t ^{\frac{1}{2}}} d t + \int 2 d t + \int t^{\frac{1}{2}} d t)

\int \frac{1}{t ^{\frac{1}{2}}} d t = 2 t^{\frac{1}{2}}

\int 2 d t = 2 t

\int t^{\frac{1}{2}} d t = \frac{2}{3} t^{\frac{3}{2}}

= \frac{3}{2} (2 t^{\frac{1}{2}} + 2 t + \frac{2}{3} t^{\frac{3}{2}})

Simplify

\frac{3}{2} (2 t^{\frac{1}{2}} + 2 t + \frac{2}{3} t^{\frac{3}{2}}) : 3 t^{\frac{1}{2}} + 3 t + t^{\frac{3}{2}}

= 3 t^{\frac{1}{2}} + 3 t + t^{\frac{3}{2}}

Add a constant to the solution

= 3 t^{\frac{1}{2}} + 3 t + t^{\frac{3}{2}} + C

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What is the integral of (3(1+t^{1/2})^2)/(2t^{1/2)} ?
The integral of (3(1+t^{1/2})^2)/(2t^{1/2)} is 3t^{1/2}+3t+t^{3/2}+C