Friday, May 4, 2012

`int (x + 4)6^((x+4)^2) dx` Find the indefinite integral

`int(x+4)6^((x+4)^2)dx=`


We will make substitution `u=(x+4)^2.` Differentiation the substitution gives us `du=2(x+4)dx.`


Now we rewrite the integral.


`1/2 int 6^((x+4)^2)2(x+4)dx=`


The above integral is equal to the starting one because `1/2` and `2` cancel out. Now we use the substitution while.


`1/2int6^udu=1/2cdot6^u/ln 6+C` 


To write the final solution we simply return the substitution i.e. we put `(x+4)^2` instead of `u.`


`6^((x+4)^2)/(2ln6)+C` where C is a constant                                                                                  

No comments:

Post a Comment