Wednesday

April 16, 2014

April 16, 2014

Posted by **John** on Thursday, May 13, 2010 at 5:03pm.

Evaluate 2x∫1 3t(t^2 + 1)^2 dt

u= t^2 + 1

du= 2t dt 1/2du=tdt

2x∫1 3u^2 1/2du

3/2 2x∫1 1/3u^3 + c

3/2 [1/3 u^3]

3/2(1/3(2x)^3) - 3/2(1/3(1)^3)

Answer= 4x - 1/2

**Related Questions**

Calculus AP - Evaluate the integral interval from [0 to pi] t sin(3t)dt Use ...

Calculus 2 correction - I just wanted to see if my answer if correct the ...

Calculus AP - I'm doing trigonometric integrals i wanted to know im doing step ...

Calculus AP - Use the table of integrals to find int cos^4 3x dx I found the ...

Calculus - How do I evaluate 2x∫1 3t(t^2 + 1)^2 dt?

calc: arc length - find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x...

COLLEGE CALCULUS. HELP! - the fundamental theorem of calculus, f(x)=∫(0,x...

Integral calculus 2 - This is also one I can't figure out: ∫ x^3 (2x^2 + ...

Calculus AP - hi again im really need help TextBook: James Stewart:Essential ...

calculus - Using an upper-case "C" for any arbitrary constants, find the general...