5.4 Indefinite Integrals and the Net Change Theorem/41: Difference between revisions
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<math>\begin{align}\int_{0}^\frac{1}\sqrt{3}\frac{t^2-1}{t^4-1} dt=\int_{0}^\frac{1}\sqrt{3} \frac{(t^2-1)}{(t^2-1)(t^2+1)} dt=\int_{0}^\frac{1}\sqrt{3} \frac{1}{(t^2+1)}dt\\[2ex]&=tan^{-1}\bigg|_{0}^{\frac{1}{\sqrt{3}}} | <math>\begin{align}\int_{0}^\frac{1}\sqrt{3}\frac{t^2-1}{t^4-1} dt&=\int_{0}^\frac{1}\sqrt{3} \frac{(t^2-1)}{(t^2-1)(t^2+1)} dt=\int_{0}^\frac{1}\sqrt{3} \frac{1}{(t^2+1)}dt\\[2ex]&=tan^{-1}\bigg|_{0}^{\frac{1}{\sqrt{3}}} | ||
\end{align}</math> | \end{align}</math> | ||
Revision as of 20:05, 1 September 2022
![{\displaystyle {\begin{aligned}\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {t^{2}-1}{t^{4}-1}}dt&=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {(t^{2}-1)}{(t^{2}-1)(t^{2}+1)}}dt=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {1}{(t^{2}+1)}}dt\\[2ex]&=tan^{-1}{\bigg |}_{0}^{\frac {1}{\sqrt {3}}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/bf027b6ecd882a6aee55f3b85cf89b92b41fceb5)