23051)Let \mathbf{t=x^{-3}}
\rightarrow\mathrm{ d}t= \frac{-3\mathrm{ d}x}{x^{4}}
\rightarrow\mathbf{I=\frac{-1}{3}\int \frac{\mathrm{d} t}{t^{2}+1}}
=\mathbf{\frac{-1}{3}tan^{-1}(t)+C}=\mathbf{\frac{-1}{3}tan^{-1}(x^{3})+C}
\hspace{-16}\bf{(1)\; \int\frac{1}{x^4.(x^6+1)}dx}$\\\\\\ $\bf{(2)\;\; \int\frac{2+\sqrt{x}}{(1+x+\sqrt{x})^2}dx}$\\\\\\ $\bf{(3)\;\; \int \left\{1+\tan x.\tan (x+\theta)\right\}dx}$\\\\\\ $\bf{(4)\;\; \int \frac{\sec x.\tan x}{3x+5}dx}$
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4 Answers
230523053)
\mathbf{I = \int 1+tan(x)\cdot tan(x+\theta)\mathrm{d}x}
=\mathbf{\int tan(x+\theta)-tan(x)\times \frac{1+tan(x)\cdot tan(x+\theta)}{tan(x+\theta )-tan(x)}}\mathrm{d}x}
=\mathbf{\int \frac{tan(x+\theta)-tan(x)}{tan(\theta)}}\mathrm{d}x}
The rest can be easily done.
