Evaluate the integral x^3/√1+x^2?

Question

Could be an indefinite integral? Could also be written as:

x^3/sqrt(1+x^2)

Many thanks!

Answer 1

put x^2+1=t



then 2xdx=dt



dy/dx=(t-1)dt/2sqrt(t)

Answer 2

The definition of a log is this: log_b a = x <---> b^x = a "b to the x-th capability provide you a" whilst the b isn't pronounced in a issue, you anticipate b to be 10. listed under are some examples: log_3 9 = 2 <---> 3² = 9 log one thousand = 3 <---> 10³ = one thousand once you spot "log x", you're telling your self "ok, log x is the exponent i'm searching for in the equation 10^? = x." or "ok, log_b x is the exponent i'm searching for in the equation b^? = x Mnemonic: "log bax" b=base, x = exponent, a = "answer" of b^x. Oops, after seeing somebody indignant's answer, i found out I forgot concerning the detrimental sign on your query. for the reason that log x is in simple terms an exponent n, -log x is -n exponent. And all of us understand that detrimental exponents bypass in the denominator.

Answer 3

∫x^3/√(1+x^2) dx =

∫(x^3+x)/√(1+x^2) dx - ∫x/√(1+x^2) dx =

∫x√(1+x^2) dx - ∫x/√(1+x^2) dx =

(1/2)∫√(1+x^2) d(1+x^2) - (1/2)∫1/√(1+x^2) d(1+x^2) =

(1/3)(√(1+x^2))^3 - √(1+x^2)+C

Answer 4

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