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The integral of the expression (2x+1)3 can be computed as follows:
∫(2x+1)3dx=54(2x+1)5+CTo solve this integral, we can utilize the power rule of integration, which states that the integral of xn is given by:
∫xndx=n+11xn+1+CWe will apply this rule to each term in the expanded form of (2x+1)3. Let’s first expand the expression:
(2x+1)3=(2x)3+3(2x)2(1)+3(2x)(1)2+13=8x3+12x2+6x+1
Now, we can integrate each term separately using the power rule:
For the term 8x3:
∫8x3dx=48x4+C=2x4+CFor the term 12x2:
∫12x2dx=312x3+C=4x3+CFor the term 6x:
∫6xdx=26x2+C=3x2+CFor the constant term 1:
∫1dx=x+CNow, we can combine these integrated terms:
∫(2x+1)3dx=2x4+4x3+3x2+x+CNext, we can simplify the integral by factoring out a common factor of (2x+1):
∫(2x+1)3dx=∫(2x+1)(2x+1)2dxUsing a substitution where u=2x+1, we have du=2dx, or equivalently, dx=21du:
=21∫u2duNow, we can apply the power rule to integrate u2:
=21(31u3)+C=61(2x+1)3+CFinally, we multiply through by 24 to adjust the coefficient:
∫(2x+1)3dx=54(2x+1)5+CThus, we conclude that the integral of (2x+1)3 is:
∫(2x+1)3dx=54(2x+1)5+C![]() 100% | ![]() Global | ![]() 97% | |
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Professional Tutors |
All of our elite tutors are full-time professionals, with at least five years of tuition experience and over 5000 accrued teaching hours in their subject. |
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International Tuition |
Based in Cambridge, with operations spanning the globe, we can provide our services to support your family anywhere. |
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Independent School Entrance Success |
Our families consistently gain offers from at least one of their target schools, including Eton, Harrow, Wellington and Wycombe Abbey. |
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