What is $f(x) = int x+5sqrt(x^2+1) dx$ if $f(2) = 7$ ?

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Answer 1 · 2017-02-20T20:58:20

$f(x)=1/2(x^2+5xsqrt(x^2+1)+5ln|x+sqrt(x^2+1)|)-98/10$

$f(x)=intx+5sqrt(x^2+1)$ $dx$

Let $x = tanu$

$dx/(du)=sec^2u$

$dx=sec^2u$ $du$

$cosu=1/sqrt(1+x^2)$

$sqrt(1+x^2)=1/cosu=secu$

$intf(x)$ $dx=intx+5intsecusec^2u$ $du$

$intx$ $dx=1/2x^2+C$

$5intsecusec^2u$ $du=5(secutanu-intsecutan^2u$ $du)$

$=5 secu tanu-5 int sec^3u$ $du+5intsec$ $du$

$10intsecusec^2u$ $du=5secutanu+5ln|secu+tanu|+C$

$5intsecusec^2u$ $du=1/2(5secutanu+5ln|secu+tanu|)+C$

$intf(x)$ $dx=1/2(x^2+5xsqrt(x^2+1)+5ln|x+sqrt(x^2+1)|)+C$

$f(2)=7=1/2(4+10sqrt(5)+5ln|2+sqrt5|)+C$

$C=-98/10$ to $2$ dp.

$f(x)=1/2(x^2+5xsqrt(x^2+1)+5ln|x+sqrt(x^2+1)|)-98/10$

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What is f(x) = int x+5sqrt(x^2+1) dxf(x) = int x+5sqrt(x^2+1) dx if f(2) = 7 f(2) = 7 ?