Find $\displaystyle \int \frac{5-4x}{\sqrt{1+2x-x^2}}\, dx$.[3] Use the substitution \(x=\tan \the

Question

Find $\displaystyle \int \frac{5-4x}{\sqrt{1+2x-x^2}}\, dx$.[3] Use the substitution $x=\tan \theta$ to show that $\int \frac{x^5}{(x^2+1)^4} \mathrm{d}x=\int \sin^m \theta \cos^n \theta \mathrm{d}\theta$, where $m$ and $n$ are integers to be determined. [2] Hence, find the exact value of $\displaystyle \int_0^{\sqrt{3}} \frac{x^5}{(x^2+1)^4} \mathrm{d}x$. [3]

Accepted Answer

$4\sqrt{1+2x-x^2}+\sin^{-1}(\dfrac{x-1}{\sqrt2})+C$; $m=5,n=1$; exact value $9/128$

2025 VJC Promo Q7

2025 VJC Promo Q7

Solution:

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