Mathematics


Evaluate:$$\displaystyle\frac { 1 } { 15 } \int \frac { d t } { 4 + t ^ { 2 } } + \frac { 4 } { 15 } \int \frac { 6 d t } { 1 + 4 t ^ { 2 } }$$


SOLUTION

$$\dfrac{1}{15}\displaystyle\int \dfrac{dt}{4+t^2}+\dfrac{4}{15}\displaystyle\int \dfrac{6dt}{1+4t^2}$$

$$=\dfrac{1}{15}\cdot \dfrac{1}{2}\tan^{-1}\left(\dfrac{t}{2}\right)+\dfrac{24}{15\times 4}\displaystyle\int \dfrac{dt}{\left(\dfrac{1}{2}\right)^2+t^2}$$

$$=\dfrac{1}{30}\tan^{-1}\left(\dfrac{t}{2}\right)+\dfrac{6}{15}\cdot\dfrac{1}{1/2}\tan^{-1}\left(\dfrac{t}{\dfrac{1}{2}}\right)+c$$

$$=\dfrac{1}{30}\tan^{-1}\left(\dfrac{t}{2}\right)+\dfrac{4}{5}\tan^{-1}(2t)+c$$.
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Subjective Medium Published on 17th 09, 2020
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