What is $f(x) = int x+3xsqrt(x^2+1) dx$ if $f(2) = 7$ ?

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Answer 1 · 2016-07-02T09:22:26

$\implies f(x) = x^2/2+(x^2+1)^{3/2} + 5(1- sqrt5)$

$f(x) = int x+3xsqrt(x^2+1) dx$

noting the pattern $D ( alpha (x^2+1)^{3/2} ) = alpha 3/2 (x^2+1)^{1/2} 2x = 3 alpha x (x^2+1)^{1/2}$ so here $alpha = 1$

$\implies f(x) = x^2/2+(x^2+1)^{3/2} +C$

$7= 2+(5)^{3/2} +C implies C = 5(1- sqrt5)$

$\implies f(x) = x^2/2+(x^2+1)^{3/2} + 5(1- sqrt5)$

What is f(x) = int x+3xsqrt(x^2+1) dxf(x) = int x+3xsqrt(x^2+1) dx if f(2) = 7 f(2) = 7 ?