Question

Calculați integralele folosind metoda integrării prin părți :

$1) \int \frac{ {x}^{2} }{ \sqrt{ {a}^{2} + {x}^{2} } } \: dx \: , x \: \in \: \mathbb{R^{*}}$

$2) \int \sqrt{ {a}^{2} + {x}^{2} } \: dx \: ,x\:\in\:\mathbb{R^{*}}$

Answer 1

$1)\displaystyle \int \dfrac{x^2}{\sqrt{a^2+x^2}} dx= \int \dfrac{x^2+a^2-a^2}{\sqrt{x^2+a^2}} dx = \int \sqrt{x^2+a^2} dx-a^2\int \dfrac{1}{\sqrt{x^2+a^2}}   \\= \int \sqrt{x^2+a^2}dx -a^2 \ln (x+\sqrt{x^2+a^2})\\ \text{Fie }I=\int \sqrt{a^2+x^2} dx \\f' = 1 , g= \sqrt {a^2+x^2}\left(\Rightarrow f=x , g'=\dfrac{x}{\sqrt{a^2+x^2}} \right) \\I=\int (x')\cdot \sqrt{a^2+x^2}dx = x\cdot \sqrt{a^2+x^2} -\int \dfrac{x^2}{\sqrt{a^2+x^2}}dx$

$\displaystyle= x\sqrt{a^2+x^2}-\int \dfrac{x^2+a^2-a^2}{\sqrt{a^2+x^2}} dx=x\sqrt{a^2+x^2} -\int \sqrt{a^2+x^2} dx +\\a^2\int \dfrac{1}{\sqrt{a^2+x^2}} dx=x\sqrt{a^2+x^2}-I+a^2\cdot \ln(x+\sqrt{x^2+a^2})   \\\text{Deci }2I=x\sqrt{a^2+x^2}+a^2\cdot \ln (x+\sqrt{x^2+a^2})\\I=\dfrac{x\sqrt{a^2+x^2}}{2}+\dfrac{a^2\cdot \ln(x+\sqrt{x^2+a^2})}{2}+C\\\text{Inlocuim si obtinem } \dfrac{x\sqrt{a^2+x^2}}{2}+\dfrac{a^2\cdot \ln(x+\sqrt{x^2+a^2})}{2}\\-{a^2\cdot \ln(x+\sqrt{x^2+a^2})=$

$\\\boxed{ \dfrac{x\sqrt{a^2+x^2}}{2}-\dfrac{a^2\cdot \ln(x+\sqrt{x^2+a^2})}{2}}\\2) \text{L-am rezolvat la 1) }.$