Question

can you help me to solve this integral and explain me?? I can t undertand :/
(x^5 + 2x^2)/ (x^6 + 2x^3+ 5 ) dx

Answer 1

ok, seems not that hard

Answer 2

\[\int\limits \frac{ x^5+2x ^{2} }{ x ^{6}+2x ^{3}+5}dx\]

Answer 3

\[\int\limits \frac{ f \prime(x) }{ f(x) }dx=\ln f(x)\]

Answer 4

wait no, his is way harder han I thought

Answer 5

can you check that it is indeed 2x squared, not x squared

Answer 6

@almavera93

Answer 7

x^6 + 2x^3+ 5=
we can write it as
\[(x^3)^2+2*x^3+1+4\]

Answer 8

\[(x^3+1)^2+4\]

Answer 9

\[\int\limits \frac{ x^5+2x^2 }{ (x^3+1)^2+4 } * dx\]

Answer 10

\[\int\limits \frac{ x^5+x^2+x^2 }{ (x^3+1)^2+4 }*dx\]

Answer 11

\[\int\limits \frac{ (x^3+1)*x^2+x^2 }{ (x^3+1)^2+4 }*dx\]

Answer 12

\[\int\limits \frac{ (x^3+1)*x^2*dx }{ (x^3+1)^2+4 } + \int\limits\frac{ x^2*dx }{ (x^3+1)^2+4} *dx\]

Answer 13

now let x^3+1=t
on differentiating
3x^2*dx=dt
x^2dx=dt/3

put these 2 in our integral

Answer 14

\[\int\limits \frac{ t*\frac{ dt }{ 3 } }{ t^2+4 }+ \int\limits \frac{ \frac{ dt }{ 3 }}{ t^2+4 }\]

Answer 15

\[\frac{ 1 }{ 3 } \int\limits \frac{ t*dt }{ t^2+4 }+\frac{ 1 }{ 3 }* \int\limits \frac{ dt }{ t^2+4}\]

Answer 16

now its simple try it now

Answer 17

@gorv 0oh thank you, that way it is easier