∫138x(2x+4)*3dx

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∫138x(2x+4)*3dx

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∫138x(2x+4)⋅3dxSimplify.Move&amp;nbsp;3&amp;nbsp;to the left of&amp;nbsp;2x+4.∫138x3⋅(2x+4)dxCancel the common factor of&amp;nbsp;8&amp;nbsp;and&amp;nbsp;2x+4.∫134x3(x+2)dxSince&amp;nbsp;43&amp;nbsp;is constant with respect to&amp;nbsp;x, move&amp;nbsp;43&amp;nbsp;out of the integral.43∫13xx+2dxDivide&amp;nbsp;x&amp;nbsp;by&amp;nbsp;x+2.43∫131-2x+2dxSplit the single integral into multiple integrals.43(∫13dx+∫13-2x+2dx)Apply the constant rule.43(x]13+∫13-2x+2dx)Since&amp;nbsp;-1&amp;nbsp;is constant with respect to&amp;nbsp;x, move&amp;nbsp;-1&amp;nbsp;out of the integral.43(x]13-∫132x+2dx)Since 22&amp;nbsp;is constant with respect to&amp;nbsp;x, move&amp;nbsp;2&amp;nbsp;out of the integral.43(x]13-(2∫131x+2dx))Multiply&amp;nbsp;2&amp;nbsp;by&amp;nbsp;-1.43(x]13-2∫131x+2dx)Let&amp;nbsp;u=x+2. Then&amp;nbsp;du=dx. Rewrite using&amp;nbsp;u&amp;nbsp;and&amp;nbsp;du.43(x]13-2∫351udu)The integral of&amp;nbsp;1u&amp;nbsp;with respect to&amp;nbsp;u&amp;nbsp;is&amp;nbsp;ln(|u|).43(x]13-2(ln(|u|)]35))Substitute and simplify.Evaluate&amp;nbsp;x&amp;nbsp;at&amp;nbsp;3&amp;nbsp;and at&amp;nbsp;1.43((3)-1-2(ln(|u|)]35))Evaluate&amp;nbsp;ln(|u|)&amp;nbsp;at&amp;nbsp;5&amp;nbsp;and at&amp;nbsp;3.43((3)-1-2((ln(|5|))-ln(|3|)))Subtract&amp;nbsp;1&amp;nbsp;from&amp;nbsp;3.43(2-2(ln(|5|)-ln(|3|)))Use the quotient property of logarithms,&amp;nbsp;logb(x)-logb(y)=logb(xy).43(2-2ln(|5||3|))Simplify.The absolute value is the distance between a number and zero. The distance between 00&amp;nbsp;and&amp;nbsp;5&amp;nbsp;is&amp;nbsp;5.43(2-2ln(5|3|))The absolute value is the distance between a number and zero. The distance between&amp;nbsp;0&amp;nbsp;and&amp;nbsp;3&amp;nbsp;is&amp;nbsp;3.43(2-2ln(53))Apply the distributive property.43⋅2+43(-2ln(53))Multiply&amp;nbsp;43⋅2.83+43(-2ln(53))Multiply&amp;nbsp;43(-2ln(53)).83+-8ln(53)3Move the negative in front of the fraction.83-8ln(53)3The result:83-8ln(53)3&amp;nbsp;&amp;nbsp;

∫138x(2x+4)*3dx

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