Quiz

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ibee2025_n29

Find $\displaystyle \int {{{x^2}\over{x^2+1}}}{\;dx}$ .
$\displaystyle {{\ln \left(x+1\right)}\over{2}}+x-{{\ln \left(x-1\right)}\over{2 }} + C$ $\displaystyle x+{{1}\over{x}} + C$ $\displaystyle {{2\,x}\over{\left(x^2+1\right)^2}} + C$ $\displaystyle x-\arctan x + C$
Find $\displaystyle \int {\tan x}{\;dx}$ .
$\displaystyle \tan x + C$ $\displaystyle \tan x-x + C$ $\displaystyle \ln \sec x + C$ $\displaystyle \ln \left(\tan x+\sec x\right) + C$
Find $\displaystyle \int {{{1}\over{x^2\,\sqrt{x^2-9}}}}{\;dx}$ by substituting $\displaystyle x=3\,\sec \theta$
$\displaystyle 27\,\int {\left(\sec \theta\right)^2\,\tan ^2\theta}{\;d \theta} = {{\left(x^2-9\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{\int {{{1}\over{\sec \theta}}}{\;d\theta}}\over{9}} = {{\sqrt{x^2-9}}\over{9\,x}} + C$ $\displaystyle 3\,\int {\left(\sec \theta\right)^2}{\;d\theta} = \sqrt{x^2-9} + C$ $\displaystyle \int {\sec \theta}{\;d\theta} = \ln \left(2\,\sqrt{x^2-9}+2\,x\right) + C$
Find $\displaystyle \int {x\,\sqrt{x^2+9}}{\;dx}$ by substituting $\displaystyle x=3\,\tan \theta$
$\displaystyle 3\,\int {\sec \theta\,\tan \theta}{\;d\theta} = \sqrt{x^2+9} + C$ $\displaystyle 27\,\int {\left(\sec \theta\right)^3\,\tan \theta}{\;d \theta} = {{\left(x^2+9\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle 3\,\int {\tan ^2\theta}{\;d\theta} = x-3\,\arctan \left({{x}\over{3}}\right) + C$ $\displaystyle {{\int {{{\sec \theta}\over{\tan ^2\theta}}}{\;d\theta} }\over{9}} = -{{\sqrt{x^2+9}}\over{9\,x}} + C$
Find $\displaystyle -\int {{{x^2}\over{x^4-1}}}{\;dx}$
$\displaystyle {{{\rm atanh}\; \left({{x}\over{2}}\right)}\over{4}}-{{\arctan \left({{x}\over{2}}\right)}\over{4}} + C$ $\displaystyle {{{\rm atanh}\; \left({{x}\over{2}}\right)}\over{16}}+{{\arctan \left({{x}\over{2}}\right)}\over{16}} + C$ $\displaystyle {{{\rm atanh}\; x}\over{2}}+{{\arctan x}\over{2}} + C$ $\displaystyle {{{\rm atanh}\; x}\over{2}}-{{\arctan x}\over{2}} + C$

Department of Mathematics
Last modified: 2026-07-16