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First tried to U sub but ended up stuck with but there remains a denominator x that I cannot sub with u in any form. $$x^2+4=u^2\\dx=\frac{udu}{x}\\$$ on my second approach, I tried trig sub but then got stuck with $\frac {\sec\theta }{1+4\tan^2\theta}d\theta$ $$x=2\tan\theta\\dx=2\sec^2\theta d\theta\\$$ now $$\int \frac{1}{\left(x^{2}+1\right)\left(\sqrt{x^{2}+4}\right)}dx\\=\int \frac{\sec^2\theta}{(4\tan^2\theta+1)\sec\theta}d\theta$$ don't know any other ways to solve this or if there is any way to proceed further

Bowei Tang
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Mahi Alam
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  • A general strategy for integrals of the type $\int\frac{\mathrm dx}{Q_1\sqrt Q_2}$ where $Q_1$ and $Q_2$ are quadratic functions in $x$ with coefficient of $x$ as $0$ can be found here. – Integreek Nov 10 '24 at 16:49

2 Answers2

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I'll start with your last integral: $$\begin{align} \int_{}^{} \frac{\sec x}{4\tan^2x+1}dx&=\int_{} \frac{\cos x}{4\sin ^2x+\cos ^2x}dx\\ &=\int \frac{1}{3\sin^2 x+1}d\sin x\\&=\int_{} \frac{\sqrt 3}{3}\frac{1}{(\sqrt 3\sin x)^2+1}d\sqrt 3\sin x\\ &=\frac{\sqrt 3}{3}\arctan (\sqrt 3\sin x)+C \end{align} $$

Bowei Tang
  • 3,763
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HINT

Another possible way to approach it:

\begin{align*} \int\frac{\mathrm{d}x}{(x^{2} + 1)\sqrt{x^{2} + 4}} & = \int\frac{\mathrm{d}x}{((x^{2} + 4) - 3)\sqrt{x^{2} + 4}}\\\\ & = \int\frac{\mathrm{d}(2\sinh(z))}{((4\sinh^{2}(z) + 4) - 3)\sqrt{4\sinh^{2}(z) + 4}}\\\\ & = \int\frac{\mathrm{d}z}{4\cosh^{2}(z) - 3}\\\\ & = \int\frac{\mathrm{d}z}{e^{2z} + e^{-2z} - 1}\\\\ & = \int\frac{e^{2z}}{e^{4z} - e^{2z} + 1}\mathrm{d}z \end{align*}

Can you take it from here?

Átila Correia
  • 18,937
  • I am still in class 12 here in india so we haven't learned hyperbolic functions yet. But I can take it from the exponential part. – Mahi Alam Jul 30 '24 at 14:58