\int\limits {3 ^{ (x)/(2) } } \, dx=

Question

`\int\limits {3 ^{ (x)/(2) } } \, dx`=

Answer 1

`\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`

Answer 2

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