Wednesday

November 26, 2014

November 26, 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

**Answer this Question**

**Related Questions**

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

Calculus - Evaluate the triple integral ∫∫∫_E (x+y)dV where E ...

Calculus - Evaluate the triple integral ∫∫∫_E (x)dV where E is...

Calculus - Evaluate the triple integral ∫∫∫_E (xy)dV where E ...

Calculus - Evaluate the triple integral ∫∫∫_E (z)dV where E is...

Calculus - Evaluate the triple integral ∫∫∫_E (xyz)dV where E ...

Calculus - Evaluate the following definite integral: integral at a = -1, b=2 -...

Integral Help - 1.) ∫ (sin x) / (cos^2 x) dx 2.) ∫ (1) / (1+x^2) dx ...

Calculus - Evaluate the triple integral ∫∫∫_E (x^2.e^y)dV ...

Calculus - Find the integral by substitution ∫ [(16 x3)/(x4 + 5)] dx &#...