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If $\frac{d}{dx}f(x)=4x^3-3/x^4$ such that $f(2)=0$. Then $f(x)$ is |
$x^3+1/x^4-129/8$ $x^4+x^3-129/8$ $x^4+1/x^3-125/8$ $x^4+1/x^3-129/8$ |
$x^4+1/x^3-129/8$ |
The anti derivative of $4x^3-3/x^4$ is $f(x)$. Hence$$f(x)=\int(4x^3-3/x^4)dx$$. Integrating this function we get $f(x)=x^4+1/x^3+C$ and substituting $f(2)=0$ we get the integration constant $C=-129/8$. Hence the result. |
