Cho hàm số $f\left( x \right)$ có ${f}'\left( x \right)=\left( x+4...

Xét $I=\int{\left( x+4 \right)\sqrt{x+1}dx},$ đặt $t=\sqrt{x+1}\Rightarrow {{t}^{2}}=x+1\Rightarrow 2tdt=dx.$ Khi đó:
$I=\int{\left( {{t}^{2}}+3 \right)t2tdx}=\int{\left( 2{{t}^{4}}+6{{t}^{2}} \right)}dx=\dfrac{2{{t}^{5}}}{5}+2{{t}^{3}}+C=\dfrac{2{{\left( x+1 \right)}^{\dfrac{5}{2}}}}{5}+2{{\left( x+1 \right)}^{\dfrac{3}{2}}}+C$
Suy ra $f\left( x \right)=\dfrac{2{{\left( x+1 \right)}^{\dfrac{5}{2}}}}{5}+2{{\left( x+1 \right)}^{\dfrac{3}{2}}}+C.$ Thay $x=0:$
$f\left( 0 \right)=\dfrac{2}{5}+2+C\Rightarrow C=-\dfrac{2}{5}.$ Do đó $f\left( x \right)=\dfrac{2}{5}{{\left( x+ \right)}^{\dfrac{5}{2}}}+2{{\left( x+1 \right)}^{\dfrac{3}{2}}}-\dfrac{2}{5}.$
Khi đó
$\int_{0}^{3}{\left( \dfrac{2}{5}{{\left( x+1 \right)}^{\dfrac{5}{2}}}+2{{\left( x+1 \right)}^{\dfrac{3}{2}}}-\dfrac{2}{5} \right)}dx=\left( \dfrac{4}{35}{{\left( x+1 \right)}^{\dfrac{7}{2}}}+\dfrac{4}{5}{{\left( x+1 \right)}^{\dfrac{5}{2}}}-\dfrac{2}{5}x \right)\left| _{\begin{smallmatrix}
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0
\end{smallmatrix}}^{\begin{smallmatrix}
3 \\

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