Solve for f(x).∫0x3+1f′(t)1+f(t)dt=ln⁡(x)

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Waylon Padilla · Accepted Answer

Step 1Notice thatf′(t)f(t)+1=[ln∘(1+f)]′(t),so the equation is equivalent to∫0x3+1[ln∘(1+f)]&#039;(t) dt=ln(x).By the fundamental theorem of calculus,∫0x3+1[ln∘(1+f)]&#039;(t) dt=ln[1+f(x3+1)]-ln[f(0)],soln[1+f(x3+1)]-ln[1+f(0)]=ln1+f(x3+1)1+f(0)=ln(x).This is equivalent to1+f(x3+1)1+f(0)=x.This is equivalent tof(x3+1)=[1+f(0)]⋅{x}−1.Let&amp;nbsp;y=x3+1, hence&amp;nbsp;x=y-13. Therefore,f(y)=[1+f(0)]·y-13-1.ince the equation requires&amp;nbsp;x&amp;gt;0, it means that&amp;nbsp;y&amp;gt;1. However, this represents a problem, as it means that 0 is not in the domain of f, and letting&amp;nbsp;y=0&amp;nbsp;implies&amp;nbsp;f(0)=−1, and we know form the original equation that&amp;nbsp;f(t)≻1.As such, the original equation has no solutions. If you replace the original integral by∫ax3+1,where&amp;nbsp;a&amp;gt;1. Accordingly, this results inf(x)=[1+f(a)]·x-13-1.One can solve for f(a), by havingf(a)=[1+f(a)]·a-13-1=a-13+a-13·f(a)-1,which is equivalent to[1-·3]a-1f(a)=a-13-1,and as long as&amp;nbsp;a≠2, this meansf(a)=a-13-11-y-13=-1,which means thatf(x)=−1which is not allowed by the original equation. So in fact, the equation can only have solutions if&amp;nbsp;a=2. f(a) then becomes a free parameter, and the familyf(x)=[1+f(2)]·x-13-1becomes an infinite family of solutions.

Landyn Whitney · Accepted Answer

Step 1⇒3x3f′(x3+1)=1+f(x3+1)⇒x(3x2)f′(x3+1)=1+f(x3+1)⇒3x2f′(x3+1)1+f(x3+1)=1xNow integrate both sides with respect x and also note that that the derivative of denominator on LHS is written in numerator, hence using substitution and some manipulation , we getf(x3+1)=Ax−1where A is a constant and thereforef(x)=A(x-1)13-1

Solve for f(x). \int_{0}^{x^3+1}\frac{f'(t)}{1+f(t)}dt=\ln(x)

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