jackpot: first of all, your question reduces to

Integrate e^x.15^e^x dx

by making the substitution u=e^x, it further reduces to

Integrate 15^u du = (1/ln15) 15^u + c = (1/ln15) 15^e^x + c.

The next time you post a simple question, I'll whoop your arsse!!!

honey: take the subtitution e^x=v, then the que. bcums §v.15^v.dx..........dv/dx= e^x, and x= in.v, making dx= dv/v, replacing this in the integral §15^v.dv= 15^v+1/(v+1), replacing v with the original values of x, we have 15^(e^x+1)/(e^x+1)

Richiez: Welcome fatty20, weldone benbuks, honey and ortarico
Woooaaaw it's like we now have our first female guru...jackpot

Calculusf(x):
...for the second question...x,7,y,15 is an a.p...x is the first term...7 is the second term...y is the third term and 15 is the fourth term...for the second term...we have already known that a=x...using the formula for a.p that a+(n-1)d...for second term.n=2... substitute that...a+d=second term...for the third term...it will be a+2d and for the third term...try yours for the fourth term...guess your answer is a+3d(great)...therefore a+d=7(second term)...a+3d=15(fourth term)...subtract the second term from the fourth term to get a+3d-(a+d)=15-7....a+3d-a-d=8...2d=8...divide through by 2...d=4...since a+d=7 then substitute for d=4...a+4=7...a=7-4=3...since x=first term(a)...therefrod x=3 and y(third term)=a+2d...y=3+2*4=3+8...y is equal to 11...therefore x=3,y=11

Sir-Tunechi:

if a,b,c,d, z an ap den a+c=2b nd b+d=2c derefore 7+15=2y, y=22/2=11 den x+y=2*7,
x+11=14
x=3. Simple

Boladearo: yes boss. Queen jackpot, welcum to d ause

benbuks: Sm ppl say. "wot a man cn do a woman cn do bdta". Shw us wot u'v got...@jacpot,.

Calculusf(x):
Integrate 7^sinx.dx

Boladearo: thats tru, prove it to us queen jackpot

jackpot: how am I doing, Sir?

jackpot: Transform "7^sin x" to "e^(sin x ln 7)".

Get the Maclaurin series expansion for the exponential function. e^x =\sum\limits_{k=1}^{\infty} \frac{x^k}{k!}

Replace "x" with "sin x ln 7"

Finally, integrate term by term.

Thats all!