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JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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The value of ∫(1+x2)dx(1−x2)1+x2+x4\int \frac{\left(1+x^2\right) d x}{\left(1-x^2\right) \sqrt{1+x^2+x^4}}, is
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Correct option: A
−123log⁡∣x4+x2+1−3xx4+x2+1+3x∣+C-\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{x^4+x^2+1}-\sqrt{3} x}{\sqrt{x^4+x^2+1}+\sqrt{3} x}\right|+C
Option A (−123log⁡∣x4+x2+1−3xx4+x2+1+3x∣+C-\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{x^4+x^2+1}-\sqrt{3} x}{\sqrt{x^4+x^2+1}+\sqrt{3} x}\right|+C) is correct because the integral simplifies to this logarithmic form after appropriate substitutions.

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∫sin⁡5xcos⁡4xdx=\int \sin ^5 x \cos ^4 x d x= If f(x)f(x) is differentibale and ∫0t2xf(x)dx=25t5\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5 then f(425)f\left(\frac{4}{25}\right) equals If ∫1−7cos⁡2xsin⁡7xcos⁡2xdx=f(x)sin⁡7x+C\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C is equal to If ∫ex2(2−1x2)de=ex2f(x)+C\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C} and f(12)=2{f}\left(\frac{1}{2}\right)=2, then f(1)f(1) is equal to (where C{C} is an arbitrary constant) ∫x7(1−x2)5dx\int \frac{x^7}{\left(1-x^2\right)^5} d x is equal to: ∫ ax3+  bx2+cx4   dx\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx equals
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