School and Work

Let x^4 = u
4 x^3 dx = du
x^3 dx = (1/4) du

Integral x^3e^x^4 dx = (1/4) integral e^u du
= e^u / 4 + C
= e^x^4 / 4 + C

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It seems impossible to integrate manually, at least to me. According to WolframAlpha, you'll end up with a gamma function in the integration.

Remember, when you substitute, you have to change the variable of respect as well. Simply substituting u = x^4 won't magically change dx to du.

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Re write the top ,i.e., x^2 as x^2+1-1

hence x^2/(1+x^2) = (x^2+1)/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2)

Hence the integral answer is x - arctan(x) +C

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integral [(x² - 1) / (x + 1)] dx

you know that x² - 1 = (x + 1) (x - 1); so let's replace it

= integral [(x - 1) (x + 1) / (x + 1)] dx; the (x + 1) on the numerator and denominator will disappear, so

= integral [x - 1] dx; then

= x²/2 - x; since we have to evaluate from x = 1 to x = 2; then

= (2)² / 2 - 2 - (1² / 2 - 1)

= 2 - 2 - (1 / 1)

=====
= -1
=====

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Home >All Categories > Science & Mathematics >Mathematics >Open Question Kushina Member since:March 26, 2011Total points:95 (Level 1)how to integrate In t cos 2t? the original equation were t^-1 sin(2t) + 4t - 1... i did let u= 1/t and dv= sin2t. so, when i used the formula uv - Integral of v du, i will get v du as In t cos 2t /2. ignore the 1/2, now i wanna integrate In t cos 2t, then i am stuck, i don't know what to do next. can someone lend a help? thank you..Answer Question

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