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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 6

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 6

Answers (1)

Answer : -        \frac{1}{\sqrt{2}} \tan ^{-1} \frac{x-1}{\sqrt{2 x}}--\frac{1}{2 \sqrt{2}} \log \left|\frac{x+1-\sqrt{2 x}}{x+1+\sqrt{2 x}}\right|+C

Hint :-                Use substitution method and special integration formula.

Given :-                 \int \frac{1}{\left(x^{2}+1\right) \sqrt{x}} d x

Sol : -             Let   \mathrm{I}=\int \frac{1}{\left(x^{2}+1\right) \sqrt{x}} d x

Put \mathrm{x}=\mathrm{t}^{2}=>\mathrm{d} \mathrm{x}=2 \mathrm{t} \mathrm{d} \mathrm{t}. then,

\begin{aligned} \mathrm{I} &=\int \frac{1}{\left(\left(t^{2}\right)^{2}+1\right) \sqrt{t^{2}}} 2 t d t \\ &=\int \frac{1}{\left(t^{4}+1\right) t} 2 t d t \end{aligned}

=2 \int \frac{1}{\left(t^{4}+1\right)} d t=2 \int \frac{\frac{1}{t^{2}}}{\frac{t^{4}+1}{t^{2}}} d t  (? dividing numerator and denominator by t2)

 

\begin{aligned} &=2 \int \frac{\frac{1}{t^{2}}}{\frac{t^{4}}{t^{2}}+\frac{1}{t^{2}}} d t=2 \int \frac{\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t \\ &=\int \frac{\frac{1}{t^{2}}+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t=\int \frac{\frac{1}{t^{2}}+1-1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}} d t \\ &=\int \frac{\left(1+\frac{1}{t^{2}}\right)+\left(\frac{1}{t^{2}}-1\right)}{t^{2}+\frac{1}{t^{2}}} d t=\int \frac{\left(1+\frac{1}{t^{2}}\right)+\left(\frac{1}{t^{2}}-1\right)}{t^{2}+\frac{1}{t^{2}}} d t \end{aligned}

\begin{aligned} &=\int \frac{\left(1+\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}} d t-\int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}} d t \\ &=\int \frac{\left(1+\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+2-2} d t-\int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+2-2} d t \\ &\begin{aligned} =\int \frac{\left(1+\frac{1}{+t^{2}}\right)}{\left.\left(t-\frac{1}{t}\right)^{2}\right)+2} d t-\int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left.\left(t+\frac{1}{t}\right)^{2}\right)-2} d t \quad \ldots \ldots \text { (I) } \quad\left(\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right.\\ &\left.(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \end{aligned} \end{aligned}
\begin{aligned} &\text { Now, } \int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left.\left(t-\frac{1}{t}\right)^{2}+2\right)} d t \\ &\text { Put t }-1 / \mathrm{t}=\mathrm{u}=>1-\left(\frac{-1}{t^{2}}\right) \mathrm{dt}=\mathrm{d} \mathrm{u} \\ &\qquad \Rightarrow\left(1+\frac{1}{t^{2}}\right) \mathrm{d} \mathrm{t}=\mathrm{d} \mathrm{u} \text { then } \\ &\qquad \therefore \int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left.\left(t-\frac{1}{t}\right)^{2}+2\right)} d t=\int \frac{d u}{\left.(u)^{2}+2\right)}=\int \frac{d u}{\left.(u)^{2}+(\sqrt{2})^{2}\right)} \end{aligned}
                                \begin{aligned} &=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+C_{1} \quad\left(\because \int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right) \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}+C_{1} \quad(\because u=t-1 / t) \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t^{2}-1}{t}+C_{1} \\ &=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{t^{2}-1}{\sqrt{2 t}}+C_{1} \ldots \ldots . . . \text { (II) } \end{aligned}

 

 

And  \int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left.\left(t+\frac{1}{t}\right)^{2}-2\right)} d t

 

            Put \mathrm{t}+1 / \mathrm{t}=\mathrm{v} \quad=>\left(1-1 / \mathrm{t}^{2}\right) \mathrm{dt}=\mathrm{d} \mathrm{y}  then,

\begin{aligned} \therefore \int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left.\left(t+\frac{1}{t}\right)^{2}-2\right)} d t &=\int \frac{d v}{\left.(v)^{2}+2\right)}=\int \frac{d v}{\left.(v)^{2}+(\sqrt{2})^{2}\right)} \\ &=\frac{1}{2 . \sqrt{2}} \log \left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C \quad\left(\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right) \\ &=\frac{1}{2 \cdot \sqrt{2}} \log \left|\frac{\left(t+\frac{1}{t}-\sqrt{2}\right)}{\left(t+\frac{1}{t}+\sqrt{2}\right)}\right|+C_{2} \end{aligned}                                                                                                                        

                                                         \begin{aligned} &=\frac{1}{2 . \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(\frac{t^{2}+1+\sqrt{2} t}{t}\right)}\right|+C_{2} \\ &=\frac{1}{2 . \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(t^{2}+1+\sqrt{2} t\right)}\right|+C_{2} \ldots \ldots . .(I I I) \end{aligned}

Put the values of equ. (II) and (III) in equ.(I), We get,

      \begin{aligned} &I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left(t^{2}-1\right)}{(\sqrt{2} t)}+C_{1}-\left(\frac{1}{2 \cdot \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(t^{2}+1+\sqrt{2} t\right)}\right|+C_{2}\right)\right. \\ &I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left(t^{2}-1\right)}{(\sqrt{2} t)}+C_{1}-\frac{1}{2 . \sqrt{2}} \log \left|\frac{\left(t^{2}+1-\sqrt{2} t\right)}{\left(t^{2}+1+\sqrt{2} t\right)}\right|-C_{2}\right) \\ &I=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{(x-1)}{(\sqrt{2} x)}-\frac{1}{2 . \sqrt{2}} \log \left|\frac{(x+1-\sqrt{2} x)}{(x+1+\sqrt{2} x)}\right|+C\right) \quad\left(\because t^{2}=x \text { and } c=c_{1}-c_{2}\right) \text { Ans.. } \end{aligned}

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