Mathematics

$$\int { \dfrac { 1 }{ { \left( 3x+1 \right)  }^{ 2 }+{ 3 }^{ 2 } } dx\quad  }$$
$$\dfrac { 1 }{ 3 } \times \dfrac { 1 }{ 3 } { tan }^{ -1 }\left( \dfrac { 3x+1 }{ 3 }  \right) +c$$


SOLUTION
$$\displaystyle\int{\dfrac{dx}{{\left(3x+1\right)}^{2}+{3}^{2}}}$$
Let $$t=3x+1\Rightarrow\,dt=3\,dx$$
$$\Rightarrow\,dx=\dfrac{1}{3}dt$$
$$\Rightarrow\,\displaystyle\int{\dfrac{dx}{{\left(3x+1\right)}^{2}+{3}^{2}}}$$
$$=\dfrac{1}{3}\displaystyle\int{\dfrac{dt}{{t}^{2}+{3}^{2}}}$$
$$=\dfrac{1}{3}\dfrac{1}{3}{\tan}^{-1}{\dfrac{t}{3}}+c$$ where $$c$$ is the constant of integration.
$$=\dfrac{1}{9}{\tan}^{-1}{\left(\dfrac{3x+1}{3}\right)}+c$$ where $$t=3x+1$$
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Subjective Medium Published on 17th 09, 2020
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