Question

exFormula1" title=" \int\limits {3 ^{ \frac{x}{2} } } \, dx " alt=" \int\limits {3 ^{ \frac{x}{2} } } \, dx " align="absmiddle" class="latex-formula">=

Answer 1

$\int 3^{\tfrac{x}{2}}\, dx=(*)\\ t=\dfrac{x}{2}\\ dt=\dfrac{x}{2}\, dx\\ dx=2\, dt\\ (*)=\int 3^t\cdot2\, dt=\\ 2\int 3^t \, dt=\\ 2\cdot\dfrac{3^t}{\ln 3}+C=\\ \boxed{\dfrac{2\cdot3^{\tfrac{x}{2}}}{\ln 3}+C}$

Answer 2

$\int { { 3 }^{ \frac { x }{ 2 } } } dx\\ \\ =\int { { \left( { 3 }^{ x } \right) }^{ \frac { 1 }{ 2 } } } dx$

However:

$u={ 3 }^{ x }\\ \\ \therefore \quad \frac { du }{ dx } ={ 3 }^{ x }\cdot \ln { 3 } \\ \\ \therefore \quad du={ 3 }^{ x }\cdot \ln { 3 } dx\\ \\ \therefore \quad dx=\frac { 1 }{ { 3 }^{ x }\cdot \ln { 3 } } du=\frac { 1 }{ u\cdot \ln { 3 } } du$

So let's use:

$\int { { u }^{ \frac { 1 }{ 2 } } } \cdot \frac { 1 }{ u\cdot \ln { 3 } } du\\ \\ =\int { \frac { 1 }{ \ln { 3 } } } \cdot { u }^{ -\frac { 1 }{ 2 } }du$

But you need to know that:

$\int { k{ u }^{ n } } du\\ \\ =\frac { k{ u }^{ n+1 } }{ n+1 } +C$

Therefore:

$\int { \frac { 1 }{ \ln { 3 } } } \cdot { u }^{ -\frac { 1 }{ 2 } }du\\ \\ =\frac { \frac { 1 }{ \ln { 3 } } \cdot { u }^{ \frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } } +C$

$\\ \\ =\frac { 1 }{ \ln { 3 } } \cdot { u }^{ \frac { 1 }{ 2 } }\cdot 2+C\\ \\ =\frac { 1 }{ \ln { 3 } } \cdot { 3 }^{ \frac { x }{ 2 } }\cdot 2+C\\ \\ =\frac { 2\cdot { 3 }^{ \frac { x }{ 2 } } }{ \ln { 3 } } +C$