Question

evaluate this integral and ill give medal

Answer 1

\[\int\limits_{}^{}(x^6+7x)^4dx\]

Answer 2

@mathmale

Answer 3

Can you just apply the power rule three times or use usub. Can you try?

Answer 4

Try these then recombine ...
\[\int\limits_{}^{} u ^{4}du\]\[u = (x ^{6}+7x)\]\[\int\limits_{}^{} x ^{6} dx + \int\limits_{}^{} 7x dx\]

Answer 5

\[\left( x^6+2x \right)^4=C _{0}^{4}(x^6)^4+C _{1}^{4}(x^6)^3(2x)+C _{2}^{4}(x^6)^2(2x)^2+C _{3}^{4}(x^6)(2x)^3+C _{4}^{4}(2x)^4\]
\[C _{0}^{4}=1,C _{1}^{4}=\frac{ 4 }{ 1 }=4,C _{2}^{4}=\frac{ 4*3 }{ 2*1 }=6,C _{3}^{4}=\frac{ 4*3*2 }{ 3*2*1 }=4,C _{4}^{4}=\frac{ 4*3*2*1 }{ 4*3*2*1 }=1\]

Answer 6

i think now you can simplify and integrate.

Answer 7

wow,what process did you just done?

Answer 8

i wrote by mistake 2x instead of 7x.
i used binomial theorem

Answer 9

you expand it right?

Answer 10

correct.

Answer 11

is there a simple process for this?