In J you have a straight forward anti derivative x+e^x
In K use u - substitution
let u = x^4 +1 the anti-derviative is then simple
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Each of these terms in the integrand is a standard form. Refer to the tables here: Lists of integrals - Wikipedia, the free encyclopedia
Make the substitution .
What's the derivative of u^n? It's n*u^(n-1)*u', right? (just chain rule, bring the exponent out front, take one from that exponent, multiply by the derivative of the inside) Like, the derivative of (x^3 + 2)^2 is 2(x^3 + 2)(3x^2).
When integrating, we just tak the opposite approach. That is to say, we raise n by 1, divide by n + 1, and then divide by the derivative of the inside.
Look again at 2(x^3 + 2)(3x^2). Let's see if we can get back our original (x^3 + 2)^2. Let u = x^3 + 2. Since we're working backward, we know the form n*u^(n-1)*u' has an integral of u^n. Our (x^3 + 2) is being raised to the power of 1, right? So when does n-1=1? When n is 2. This means that so far, we're in the correct form n*u^(n-1). However, we haven't accounted for u'. If we are infact multiplying n*u^(n-1) by u', than we know our answer is just u^n. Lucky us, (3x^2) is u'. So it's in the correct form to convert to u^n, where n is 2 and u is x^3 + 2.
Same concept with x^3(x^4 + 1)^(1/2). Let u=x^4 + 1. Our n is then 1/2 + 1 (or 3/2). Next, however, we don't have a 3/2 coefficient, do we? We have a coefficient of 1. So, multiply by 2/3 to account for that. There's just one other thing to check before we convert to u^n, our u'. Unfortunately again, we're multiplying by x^3, not 4x^3. But, like before, we just multiply by 1/4 to balance it out.
Thus you get your answer of (2/3)(1/4)u^(3/2) or (1/6)u^(3/2)