MHT CET 2025 21st April Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& 5𝑃(𝑋=0)=𝑃(𝑋=1)\\ & \Rightarrow 5^{10}{𝐶}_0\left(𝑝{)}^0\right(𝑞{)}^{10-0}={}^{10}{𝐶}_1\left(𝑝{)}^1\right(𝑞{)}^{10-1}\\ & \Rightarrow 5{𝑞}^{10}=10{𝑝}^9\\ & \Rightarrow 𝑞=2𝑝\\ & \Rightarrow 1-𝑝=2𝑝\dots [𝑝+𝑞=1]\\ & \Rightarrow 1=3𝑝\\ & \Rightarrow 𝑝=\frac{1}{3}\\ & \Rightarrow 𝑞=1-𝑝=1-\frac{1}{3}=\frac{2}{3}\\ & \begin{array}{cc}\therefore \frac{𝑃(𝑋=5)}{𝑃(𝑋=6)}& =\frac{{}^{10}{𝐶}_5{\left(\frac{1}{3}\right)}^5{\left(\frac{2}{3}\right)}^5}{{}^{10}{𝐶}_6{\left(\frac{1}{3}\right)}^6{\left(\frac{2}{3}\right)}^4}\\ & =\frac{6}{5}\times 2\\ & =\frac{12}{5}\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \frac{d𝑦}{d𝑥}=(𝑥+9𝑦{)}^2\\ & \text{ Let }u=𝑥+9𝑦\\ \therefore & \frac{\mathrm{du}}{d𝑥}=1+9\frac{d𝑦}{d𝑥}\\ & \Rightarrow \frac{\mathrm{du}}{d𝑥}=1+9u^2\\ & \Rightarrow \frac{1}{1+9u^2}\mathrm{du}=d𝑥\end{array}\\ & \text{ Integrating on both sides, we get }\\ & \int \frac{1}{1+9{𝑢}^2}𝑑𝑢=\int 𝑑𝑥\end{array}$

$\begin{array}{cc}& \Rightarrow \frac{1}{3}{\tan }^{-1}\left(3u\right)=𝑥+c\\ & \Rightarrow {\tan }^{-1}\left(3\right(𝑥+9𝑦\left)\right)=3𝑥+3c\\ & \Rightarrow {\tan }^{-1}(3𝑥+27𝑦)=3𝑥+3c\dots \text{ (i) }\\ & \text{ When }𝑥=0,𝑦=\frac{1}{27}\\ \therefore & {\tan }^{-1}\left(1\right)=0+3c\\ & \Rightarrow 3c=\frac{𝜋}{4}\\ \therefore & {\tan }^{-1}(3𝑥+27𝑦)=3𝑥+\frac{𝜋}{4}\dots \left[\text{ From (i) }\right]\\ & \Rightarrow 3𝑥+27𝑦=\tan \left(3(𝑥+\frac{𝜋}{12})\right)\end{array}$

$\begin{array}{cc}\underset{𝑥\to \infty }{lim}\frac{e^{{𝑥}^4}-1}{e^{{𝑥}^4}+1}& =\underset{𝑥\to \infty }{lim}\frac{(1-\frac{1}{e^{{𝑥}^4}})}{(1+\frac{1}{e^{{𝑥}^4}})}\\ & =\frac{\underset{𝑥\to \infty }{lim}(1-\frac{1}{e^{{𝑥}^4}})}{\underset{𝑥\to \infty }{lim}(1+\frac{1}{e^{{𝑥}^4}})}\\ & =\frac{1-0}{1+0}\\ & =1\end{array}$

$\begin{array}{cc}& \text{ Here, }\overline{{𝑎}_1}=4\widehat{𝑖}-\widehat{𝑗},\overline{{𝑏}_1}=\widehat{𝑖}+2\widehat{𝑗}-3\widehat{𝑘}\\ & \overline{{𝑎}_2}=\widehat{𝑖}-\widehat{𝑗}+2\widehat{𝑘},{\overline{𝑏}}_2=2\widehat{𝑖}+4\widehat{𝑗}-5\widehat{𝑘}\\ & \begin{array}{cc}{\overline{𝑎}}_2-{\overline{𝑎}}_1& =-3\widehat{𝑖}+2\widehat{𝑘}\\ {\overline{𝑏}}_1\times {\overline{𝑏}}_2=& \mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 1& 2& -3\\ 2& 4& -5\end{array}\mid =2\widehat{𝑖}-\widehat{𝑗}\end{array}\\ & \begin{array}{cc}\mid {\overline{𝑏}}_1\times {\overline{𝑏}}_2\mid & =\sqrt{\surd 5}\end{array}\\ & \text{ Shortest distance }=\mid \frac{({\overline{𝑎}}_2-{\overline{𝑎}}_1)\cdot ({\overline{𝑏}}_1\times {\overline{𝑏}}_2)\mid }{\mid {\overline{𝑏}}_1\times {\overline{𝑏}}_2\mid }\mid \\ & =\mid \frac{(-3\widehat{𝑖}+2\widehat{𝑘})\cdot (2\widehat{𝑖}-\widehat{𝑗})\mid }{\sqrt{\surd 5}}\mid \\ & =\frac{6}{\sqrt{\surd 5}}\text{ units }\end{array}$

$\begin{array}{cc}& f\left({𝑥}^2\right)f\left({𝑦}^2\right)-\frac{1}{2}(f\left(\frac{{𝑥}^2}{{𝑦}^2}\right)+f\left({𝑥}^2{𝑦}^2\right))\\ & \begin{array}{c}=\cos (\log {𝑥}^2)\cdot \cos (\log {𝑦}^2)-\frac{1}{2}[\cos [\log \left(\frac{{𝑥}^2}{{𝑦}^2}\right)]+\cos [\log \left({𝑥}^2{𝑦}^2\right)]]\end{array}\\ & \begin{array}{c}=\cos (\log {𝑥}^2)\cdot \cos (\log {𝑦}^2)12[\cos (\log {𝑥}^2-\log {𝑦}^2)+\cos (\log {𝑥}^2+\log {𝑦}^2)]\\ =\cos (\log {𝑥}^2)\cdot \cos (\log {𝑦}^2)-\frac{1}{2}[2\cos (\log {𝑥}^2)\cdot \cos (\log {𝑦}^2)]\end{array}\\ & =0\end{array}$

$\begin{array}{cc}& \frac{d𝑦}{d𝑥}=2𝑥𝑦e^{{𝑥}^2}\\ & \Rightarrow \frac{1}{𝑦}d𝑦=2𝑥e^{{𝑥}^2}d𝑥\end{array}$

$\begin{array}{cc}& \text{ Integrating on both sides, we get }\\ & \begin{array}{cc}& \int \frac{1}{𝑦}d𝑦=\int 2𝑥e^{{𝑥}^2}d𝑥\\ & \text{ Put }{𝑥}^2=u\Rightarrow 2𝑥d𝑥=\mathrm{du}\\ \therefore & \log 𝑦=\int e^u\mathrm{du}\\ \Rightarrow & \log 𝑦=e^u+\log c\\ \Rightarrow & \log 𝑦-\log c=e^u\\ \Rightarrow & \log \left(\frac{𝑦}{c}\right)=e^u\\ \Rightarrow & \frac{𝑦}{c}=e^{e^u}\\ \Rightarrow & 𝑦=c\cdot e^u\\ \Rightarrow & 𝑦=c\cdot e^{e^{{𝑥}^2}}\end{array}\end{array}$

A function $𝑓(𝑥,𝑦)$ is called homogeneous of degree $𝑛$ if for all real numbers $𝑘$,

$𝑓(𝑘𝑥,𝑘𝑦)={𝑘}^{𝑛}𝑓(𝑥,𝑦)$

Let's check each option:

Option A: ${𝑦}^2+2𝑥𝑦$

Replace $𝑥$ by $𝑘𝑥$ and $𝑦$ by $𝑘𝑦$:

$𝑓(𝑘𝑥,𝑘𝑦)=(𝑘𝑦{)}^2+2(𝑘𝑥\left)\right(𝑘𝑦)={𝑘}^2{𝑦}^2+2{𝑘}^2𝑥𝑦={𝑘}^2({𝑦}^2+2𝑥𝑦)$

So, it's homogeneous of degree $2$.

Option B: $2𝑥-3𝑦$

$𝑓(𝑘𝑥,𝑘𝑦)=2\left(𝑘𝑥\right)-3\left(𝑘𝑦\right)=𝑘(2𝑥-3𝑦)$

So, it's homogeneous of degree $1$.

Option C: $\sin \left(\frac{𝑦}{𝑥}\right)$

$𝑓(𝑘𝑥,𝑘𝑦)=\sin \left(\frac{𝑘𝑦}{𝑘𝑥}\right)=\sin \left(\frac{𝑦}{𝑥}\right)$

So, $𝑓(𝑘𝑥,𝑘𝑦)=𝑓(𝑥,𝑦)$. This is homogeneous of degree $0$.

Option D: $\cos 𝑥+\sin 𝑦$

$𝑓(𝑘𝑥,𝑘𝑦)=\cos \left(𝑘𝑥\right)+\sin \left(𝑘𝑦\right)$

This is not a homogeneous function because it does not satisfy $𝑓(𝑘𝑥,𝑘𝑦)={𝑘}^{𝑛}𝑓(𝑥,𝑦)$ for any $𝑛$.

Final Answer:

Option D: $\overline{)\cos 𝑥+\sin 𝑦}$ is not a homogeneous function.

Let $\bar{a}=\hat{i}+𝑥\hat{j}+\hat{k}$,

$\begin{array}{cc}& \bar{b}=\hat{j}+𝑥\hat{k}\\ & \bar{c}=𝑥\hat{i}+\hat{k}\end{array}$

Volume of parallelopiped

$\begin{array}{cc}& =\left[\begin{array}{ccc}\overline{𝑎}& \overline{𝑏}& \overline{𝑐}\end{array}\right]\\ & =\mid \begin{array}{ccc}1& 𝑥& 1\\ 0& 1& 𝑥\\ 𝑥& 0& 1\end{array}\mid \\ & =1(1-0)-𝑥(0-{𝑥}^2)+1(0-𝑥)\\ & =1+{𝑥}^3-𝑥\\ & \text{ Let }f\left(𝑥\right)=1+{𝑥}^3-𝑥...\left(𝑖\right)\\ \therefore & f^{\prime }\left(𝑥\right)=3{𝑥}^2-1\end{array}$

Consider, $f^{\prime }\left(𝑥\right)=0$

$\begin{array}{cc}& \Rightarrow 3{𝑥}^2-1=0\\ & \Rightarrow 3{𝑥}^2=1\\ & \Rightarrow {𝑥}^2=\frac{1}{3}\\ & \Rightarrow 𝑥=\pm \frac{1}{\sqrt{\surd 3}}\end{array}$

Also, $f^{\prime \prime }\left(𝑥\right)=6𝑥$

When $𝑥=\frac{1}{\sqrt{\surd 3}},f^{\prime \prime }\left(𝑥\right)=\frac{6}{\sqrt{\surd 3}}>0$

$\therefore f\left(𝑥\right)$ is minimum at $𝑥=\frac{1}{\sqrt{\surd 3}}$.

When $𝑥=\frac{-1}{\sqrt{\surd 3}},f^{\prime \prime }\left(𝑥\right)=\frac{-6}{\sqrt{\surd 3}}<0$

$\therefore f\left(𝑥\right)$ is maximum at $𝑥=\frac{-1}{\sqrt{\surd 3}}$.

$\therefore$ Maximum value $=1+{\left(\frac{-1}{\sqrt{\surd 3}}\right)}^3-\left(\frac{-1}{\sqrt{\surd 3}}\right)\dots \left[\text{ From (i) }\right]$

$\begin{array}{cc}& \Rightarrow \text{ Max. value }=1+\frac{2}{3\sqrt{\surd 3}}\\ & \therefore \text{ Minimum value }=1+{\left(\frac{1}{\sqrt{\surd 3}}\right)}^3-\left(\frac{1}{\sqrt{\surd 3}}\right)\\ & \Rightarrow \text{ Min. value }=1-\frac{2}{3\sqrt{\surd 3}}\dots [\text{ From (i)] }\end{array}$

$\begin{array}{cc}& \text{ Since }𝑃\left(𝑋\right)\ge 0\\ \therefore & \frac{1+𝑝}{5}\ge 0\Rightarrow 𝑝\ge -1\\ & \frac{2-2𝑝}{5}\ge 0\Rightarrow 𝑝\le 1\\ & \frac{2-𝑝}{5}\ge 0\Rightarrow 𝑝\le 2\\ & \frac{2𝑝}{5}\ge 0\Rightarrow 𝑝\ge 0\\ \therefore & 0\le 𝑝\le 1\end{array}$

$\begin{array}{cc}& \therefore \text{ The minimum value of }𝑝\text{ is }0\text{. }\\ & \begin{array}{cc}E\left(X\right)& =\sum {𝑥}_i\cdot P\left({𝑥}_i\right)\\ & =0+1\left(\frac{2-2p}{5}\right)+2\left(\frac{2-p}{5}\right)+3\left(\frac{2p}{5}\right)\\ & =\frac{2-2𝑝+4-2𝑝+6𝑝}{5}\\ \therefore E\left(X\right)& =\frac{6+2𝑝}{5}\\ \therefore 5E\left(X\right)& =6+2𝑝\\ & \text{ When, }𝑝=0\\ 5E\left(X\right)& =6+2\left(0\right)=6\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }(\bar{a}\times (\bar{b}\times \bar{c}\left)\right)\cdot (\bar{a}\times \bar{c})=5\\ & \Rightarrow \left(\right(\bar{a}\cdot \bar{c})\bar{b}-(\bar{a}\cdot \bar{b}\left)\bar{c}\right)\cdot (\bar{a}\times \bar{c})=5\\ & \Rightarrow (\bar{a}\cdot \bar{c})(\bar{b}\cdot (\bar{a}\times \bar{c}\left)\right)-(\bar{a}\cdot \bar{b})(\bar{c}\cdot (\bar{a}\times \bar{c}\left)\right)=5\\ & \Rightarrow (\bar{a}\cdot \bar{c})(\bar{b}\cdot (\bar{a}\times \bar{c}\left)\right)-0=5\\ & \Rightarrow (\bar{a}\cdot \bar{c})(\bar{b}\cdot (\bar{a}\times \bar{c}\left)\right)=5...\left(𝑖\right)\end{array}$

Consider,

$\begin{array}{cc}& \begin{array}{cc}\overline{𝑎}\cdot \overline{𝑐}& =\left|𝑎\right|\cdot \left|\overline{𝑐}\right|\cos 𝜃\\ & \\ & =1\cdot 1\cos \frac{𝜋}{3}=\frac{1}{2}\end{array}\\ \therefore & \frac{1}{2}(\overline{𝑏}\cdot (\overline{𝑎}\times \overline{𝑐}\left)\right)=5& \dots \left[\text{ From (i) }\right]\\ & \Rightarrow \overline{𝑏}\cdot (\overline{𝑎}\times \overline{𝑐})=10& \\ & \Rightarrow \left[\overline{𝑎}\overline{𝑏}\overline{𝑐}\right]=-10& \dots [\because [𝑎b𝑐]=-[𝑏a𝑐\left]\right]\end{array}$

$\begin{array}{cc}& \sum _{𝑥=0}^{\infty }P(X=𝑥)=1\\ & \Rightarrow \sum _{𝑥=0}^{\infty }k(𝑥+1){\left(\frac{1}{2}\right)}^{𝑥}=1\\ & \Rightarrow k[1+2\left(\frac{1}{2}\right)+3{\left(\frac{1}{2}\right)}^2+\dots ]=1\\ & \Rightarrow k{[\frac{1}{1-\frac{1}{2}}+\frac{1\times \frac{1}{2}}{{(1-\frac{1}{2})}^2}]}^{=1}=1\\ & \dots \left[\begin{array}{c}\because a+(a+d)r+(a+2d)r^2+\dots \\ =\frac{a}{1-r}+\frac{\mathrm{dr}}{1-r^2}\end{array}\right]\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑘(2+2)=1\\ & \Rightarrow 𝑘=\frac{1}{4}\end{array}$

$\begin{array}{cc}\text{ Required area }& =2{\int }_{4𝑎}^{𝑎}\sqrt{\surd 4𝑎𝑥}𝑑𝑥\\ & =4\sqrt{\surd 𝑎}{\int }_{4𝑎}^{𝑎}\sqrt{\surd 𝑥}𝑑𝑥\\ & =4\sqrt{\surd 𝑎}{\left[\frac{2}{3}𝑥\sqrt{\surd 𝑥}\right]}_{4𝑎}^{𝑎}\\ & =\frac{8}{3}\sqrt{\surd 𝑎}[4𝑎\sqrt{\surd 4𝑎}-𝑎\sqrt{\surd 𝑎}]\\ & =\frac{8}{3}\sqrt{\surd 𝑎}[8𝑎\sqrt{\surd 𝑎}-𝑎\sqrt{\surd 𝑎}]\\ & =\frac{8}{3}\sqrt{\surd 𝑎}\left(7𝑎\sqrt{\surd 𝑎}\right)\\ & =\frac{56}{3}{𝑎}^2\text{ sq. units }\end{array}$

$\text{ Let }I={\int }_4^2\frac{\log {𝑥}^2}{\log {𝑥}^2+\log (36-12𝑥+{𝑥}^2)}d𝑥...\left(𝑖\right)$

$\begin{array}{c}\therefore 𝐼={\int }_4^2\frac{\log (6-𝑥{)}^2}{\log (6-𝑥{)}^2+\log {𝑥}^2}d𝑥...\left(𝑖𝑖\right)\\ [\because {\int }_b^{a}f(𝑥)d𝑥={\int }_b^{a}f(a+b-𝑥\left)\mathrm{dx}\right]\end{array}$

Adding (i) and (ii), we get

$\begin{array}{cc}& 2I={\int }_4^{2}1d𝑥=[𝑥{]}_4^2=4-2=2\\ \therefore & I=1\end{array}$

Alternate Method:

Here, $f\left(𝑥\right)=\log {𝑥}^2,a=2$ and $b=4$

$\begin{array}{c}\therefore {\int }_4^2\frac{\log {𝑥}^2}{\log {𝑥}^2+\log (36-12𝑥+{𝑥}^2)}d𝑥=\frac{1}{2}(4-2)=1\\ \dots [\because {\int }_b^a\frac{f\left(𝑥\right)}{f\left(𝑥\right)+f(a+b-𝑥)}d𝑥=\frac{1}{2}(b-a\left)\right]\end{array}$

Given, $\overline{𝑎}\cdot \overline{𝑏}=\overline{𝑎}\cdot \overline{𝑐}=0$

$\Rightarrow$ Unit vector $\bar{a}$ is perpendicular to both $\bar{b}$ and $\bar{c}$

$\begin{array}{cc}\therefore & \overline{𝑎}=\pm \frac{\overline{𝑏}\times \overline{𝑐}}{|\overline{𝑏}\times \overline{𝑐}|}\\ & |\overline{𝑏}\times \overline{𝑐}|=\left|\overline{𝑏}\right|\left|\overline{𝑐}\right|\sin 𝜃\end{array}$

$\begin{array}{c}=1\cdot 1\cdot \sin \frac{𝜋}{6}=\frac{1}{2}\\ \therefore \overline{𝑎}=\pm \frac{\overline{𝑏}\times \overline{𝑐}}{\frac{1}{2}}\\ \therefore \overline{𝑎}=\pm 2(\overline{𝑏}\times \overline{𝑐})\end{array}$

$\begin{array}{cc}& |\bar{a}+\bar{b}{|}^2=|\bar{a}{|}^2+|\bar{b}{|}^2+2\bar{a}\cdot \bar{b}\\ & \Rightarrow (\sqrt{\surd 3}{)}^2=(1{)}^2+(1{)}^2+2\bar{a}\cdot \bar{b}\\ & \Rightarrow 3=2+2\bar{a}\cdot \bar{b}\\ & \Rightarrow 2\bar{a}\cdot \bar{b}=1\\ & \Rightarrow \bar{a}\cdot \bar{b}=\frac{1}{2}\\ & \bar{a}\cdot \bar{b}=\left|\bar{a}\right|\left|\bar{b}\right|\cos 𝜃\\ & \Rightarrow \frac{1}{2}=1\cdot 1\cdot \cos 𝜃\\ & \Rightarrow \frac{1}{2}=\cos 𝜃\\ & \Rightarrow 𝜃=\frac{𝜋}{3}\end{array}$

$\begin{array}{cc}& {\int }_2^0\left[{𝑥}^2\right]d𝑥\\ & ={\int }_1^0\left[{𝑥}^2\right]d𝑥+{\int }_{\sqrt{\surd 2}}^1\left[{𝑥}^2\right]d𝑥+{\int }_{\sqrt{\surd 3}}^{\sqrt{\surd 2}}\left[{𝑥}^2\right]d𝑥+{\int }_2^{\sqrt{\surd 3}}\left[{𝑥}^2\right]d𝑥\\ & ={\int }_1^{0}0d𝑥+{\int }_{\sqrt{\surd 2}}^{1}1d𝑥+{\int }_{\sqrt{\surd 3}}^{\sqrt{\surd 2}}2d𝑥+{\int }_{\sqrt{\surd 2}}^{\sqrt{\surd 3}}3d𝑥\\ & =0+1(\sqrt{\surd 2}-1)+2(\sqrt{\surd 3}-\sqrt{\surd 2})+3(2-\sqrt{\surd 3})\\ & =\sqrt{\surd 2}-1+2\sqrt{\surd 3}-2\sqrt{\surd 2}+6-3\sqrt{\surd 3}\\ & =5-\sqrt{\surd 2}-\sqrt{\surd 3}\end{array}$

Let $I=\int \frac{3\sin 𝑥\cdot \cos 𝑥}{4\sin 𝑥+7}d𝑥$

Put $4\sin 𝑥+7=u\Rightarrow \sin 𝑥=\frac{u-7}{4}$

Differentiating w.r.t. $𝑥$, we get

$\begin{array}{cc}& 4\cos 𝑥𝑑𝑥=𝑑𝑢\Rightarrow \cos 𝑥𝑑𝑥=\frac{𝑑𝑢}{4}\\ \therefore & 𝐼=\int \frac{3\left(\frac{𝑢-7}{4}\right)\cdot \frac{𝑑𝑢}{4}}{𝑢}\end{array}$

$\begin{array}{cc}& =\frac{3}{16}\int \frac{u-7}{u}\mathrm{du}\\ & =\frac{3}{16}\int (1-\frac{7}{u})\mathrm{du}\\ & =\frac{3}{16}(u-7\log u)+c_1\\ & =\frac{3}{16}u-\frac{21}{16}\log u+c_1\\ & =\frac{3}{16}(4\sin 𝑥+7)-\frac{21}{16}\log (4\sin 𝑥+7)+c_1\\ & =\frac{12}{16}\sin 𝑥+\frac{21}{16}-\frac{21}{16}\log (4\sin 𝑥+7)+c_1\\ & =\frac{3}{4}\sin 𝑥-\frac{21}{16}\log (4\sin 𝑥+7)+c,\end{array}$

where $c=c_1+\frac{21}{16}$

Comparing with the given expression, we get

$A=\frac{3}{4},B=\frac{21}{16}$

$\therefore 𝐴+𝐵=\frac{33}{16}$

$\begin{array}{cc}& \text{ Let }\overline{𝑎}=\widehat{𝑖}-\widehat{𝑘},\overline{𝑏}=𝜆\widehat{𝑖}+\widehat{𝑗}+(1-𝜆)\widehat{𝑘}\\ & \overline{𝑐}=𝜇\widehat{𝑖}+𝜆\widehat{𝑗}+(1+𝜆-𝜇)\widehat{𝑘}\end{array}$

Volume of parallelopiped

$\begin{array}{cc}& =\left[\begin{array}{ccc}\overline{𝑎}& \overline{𝑏}& \overline{𝑐}\end{array}\right]\\ & =\mid \begin{array}{ccc}1& 0& -1\\ 𝜆& 1& 1-𝜆\\ 𝜇& 𝜆& 1+𝜆-𝜇\end{array}\mid \end{array}$

Applying $C_3\to C_3+C_1$, we get

Volume of parallelopiped

$=\mid \begin{array}{ccc}1& 0& 0\\ 𝜆& 1& 1\\ 𝜇& 𝜆& 1+𝜆\end{array}\mid$

$=1$, which is independent of both $𝜆$ and $𝜇$.

$\begin{array}{cc}& \begin{array}{cc}& \text{ Let }𝐼=\int \frac{𝑑𝑥}{𝑥({𝑥}^2+1)}\\ & \text{ Let }\frac{1}{𝑥({𝑥}^2+1)}=\frac{𝐴}{𝑥}+\frac{𝐵𝑥+𝐶}{{𝑥}^2+1}\\ \therefore & 1=𝐴({𝑥}^2+1)+(𝐵𝑥+𝐶)\cdot 𝑥\\ \Rightarrow & 1=𝐴{𝑥}^2+𝐴+𝐵{𝑥}^2+𝐶𝑥\\ \Rightarrow & 1=(𝐴+𝐵){𝑥}^2+𝐶𝑥+𝐴\end{array}\\ & \text{ Comparing the coefficients, we get }\\ & \begin{array}{cc}𝐴& +𝐵=0,𝐶=0,𝐴=1\\ \Rightarrow & 𝐵=-1,𝐶=0,𝐴=1\\ \therefore 𝐼& =\int (\frac{1}{𝑥}+\frac{-𝑥}{{𝑥}^2+1})𝑑𝑥\\ & =\int \frac{1}{𝑥}𝑑𝑥-\frac{1}{2}\int \frac{2𝑥}{{𝑥}^2+1}𝑑𝑥\\ & =\log 𝑥-\frac{1}{2}\log ({𝑥}^2+1)+𝑐\end{array}\end{array}$