mathematics-class-12-unit-07-chapter-11-indefinite-integral-l-4-7-6p-jxdg2ssw

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- $ I  =\int \frac{d x}{x^2+a^2} $

- Solution
- $ x=a \tan t \quad \Rightarrow d x=a \sec ^2 t d t $

- $ x^2+a^2 =a^2 \tan ^2 t+a^2= a^2(\tan ^2 t+1)=a^2 \sec ^2 t $

- $ =\int \frac{a \sec ^2 t d t}{a^2 \operatorname{sec}^2 t} $

- $ =\frac{1}{a} \int d t=\frac{1}{a} t+c=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c $

- $ \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c $

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<footer style = "font-size: 18px"> $\rightarrow$ Indefinite integral lecture-11 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- [$ \int \sec t d t =\log |\sec t+\tan t|+c $ ]

- $ =\log |\sec t+\tan t|+c $

- $ [ sec ^2 t=1+\tan ^2 t ] $

- $ [ \sec t =\sqrt{1+\tan ^2 t}] $

- $ =\log |\sqrt{1+\tan ^2 t}+\tan t| + c $

- $ =\log \left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|+C $

- $ =\log \left|x+\sqrt{x^2+a^2}\right|-\log |a|+c $

- $ \int \frac{d x}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C $ 
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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- $ I =\int \frac{d x}{x^2-a^2} $

- Solution:
- $ [ A^2-B^2=(A-B)(A+B) ] $

- $ =\int \frac{d x}{(x-a)(x+a)} $

- $\\{\because \frac{1}{(x-a)(x+a)}=\frac{1}{2 a}\left[\frac{2 a}{(x+a)(x-a)}\right] =\frac{1}{2a}\left[\frac{(x+a)-(x-a)}{(x+a)(x-a)}\right] =\frac{1}{2 a}\left[\frac{1}{x-a}-\frac{1}{x+a}\right]\\} $

- $ =\frac{1}{2 a} \int\left[\frac{1}{x-a}-\frac{1}{x+a}\right] d x $

- $ =\frac{1}{2 a} \int \frac{d x}{x-a}-\frac{1}{2 a} \int \frac{d x}{x+a} $

- $ =\frac{1}{2 a} \log |x-a|-\frac{1}{2 a} \log |x+a|+C $

- $ I=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C $

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Problem-4 $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- $ \int \frac{d x}{a^2-x^2} $

- solution:
- $ =\int \frac{1}{2 a}\left[\frac{(a-x)+(a+x)}{(a-x)(a+x)}\right] d x $

- $ =\frac{1}{2 a}\left[\int \frac{d x}{a+x}+\int \frac{d x}{a-x}\right] \quad\left[\int \frac{d x}{a-x}=?\right] $

- $ =\frac{1}{2 a}\left[\log |a+x|+\frac{\log |a-x|}{-1}\right]+c $

- $ =\frac{1}{2 a}[\log |a+x|-\log |a-x|]+c $

- $ \int \frac{d x}{a^2-x^2}  =\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C $ 
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Problem-5 $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- $ I=\int \frac{3 x^2 d x}{x^6+4} $

- Solution:
- $ =\int \frac{3 x^2 d x}{\left(x^3\right)^2+2^2} [\quad x^3=t \rightarrow\quad 3 x^2 d x=d t ] $

- $ =\int \frac{d t}{t^2+2^2}=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C $

- $[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a}. \tan ^{-1} \frac{x}{a}+c ] $

- $ =\frac{1}{2} \tan ^{-1} \frac{x^3}{2}+c $

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Problem-6 $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- $ \int \frac{x^2 d x}{1-x^6} $
- $ =\int \frac{x^2 d x}{1-\left(x^3\right)^2} $
- $ [ x^3=t ]$

- $ [ 3 x^2 d x=d t ] $

- $[ x^2 d x=\frac{d t}{3} ] $
- $  =\int \frac{1}{1-t^2} \frac{d t}{3} $
- $ =\frac{1}{3} \int \frac{d t}{1-t^2} $

- $ [ \int \frac{d x}{a^2-x^2}= \frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c ] $

- $ =\frac{1}{3 \times 2} \log \left|\frac{1+t}{1-t}\right|+C $

- $=\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|+C $
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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Problem-7 $\rightarrow$ Problem-8 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ =\log |\sec t+\tan t|+C $

- [x = $ a \sec t $]

- $ [ \ sec t = \frac{x}{a} ]$

- $ [ tant = \sqrt{\sec ^2 t-1} =\sqrt{\frac{x^2}{a^2} -1} ] $

- $ =\log \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|+c $

- $ \int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+c $ 
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<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Problem-8 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-9 $\rightarrow$ Problem-10 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-9 </primary>
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- $  I =\int \frac{d x}{ax^2+b x+c} $

- Solution:
- $ a x^2+b x+c =a\left(x^2+\frac{b}{a} x+\frac{c}{a}\right) $

- $ =a(\underbrace{x^2+2 \times \frac{b}{2 a} \times x+\frac{b^2}{4 a^2}}-\frac{b^2}{4 a^2}+\frac{c}{a}) $

- $ =a\left[\left(x+\frac{b}{2 a}\right)^2+\left(\frac{c}{a}-\frac{b^2}{4 a^2}\right)\right] $

- $ x+\frac{b}{2 a}=X, \quad \frac{c}{a}-\frac{b^2}{4 a^2}= \pm k^2, d x=d X $

- $ I=\frac{1}{a} \int \frac{d X}{X^2 \pm k^2} $

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<footer style = "font-size: 18px">Problem-8 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-9 </span> $\rightarrow$ Problem-10 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-indefinite-integral-l-4-success"><Success style="font-size:25px"> Indefinite Integral L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$ I =\int \frac{d x}{9\left((x+1 / 3)^2+(2 / 3)^2\right)}=\frac{1}{9} \int \frac{d X}{X^2+(2 / 3)^2} $</p>
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<p>Solution:</p>
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<p>(Put $x+ \frac{1}{3} = X$)</p>
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<p>$=\frac{1}{9} \times \frac{1}{2 / 3} \tan ^{-1} \frac{X}{2 / 3}+C$</p>
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<p>$=\frac{1}{6} \tan ^{-1} \frac{3}{2}\left(x+\frac{1}{3}\right)+C $</p>
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<p>$=\frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+C  $</p>
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<footer style = "font-size: 18px">Problem-9 $\rightarrow$ Problem-10 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-11 $\rightarrow$ Problem-12 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-11 </primary>
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- $ I=\int \frac{d x}{9 x^2+6 x+5} $

- Solution:
- $ 9 x^2+6 x+5=(3 x)^2+2(3 x)+1+4=(3 x+1)^2+4 $

- $ I =\int \frac{d x}{(3 x+1)^2+4} $

- [ 3 x+1=t ]

- [ 3 d x=d t ]

- $ =\frac{1}{3} \int \frac{d t}{t^2+2^2}=\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1}\frac{t}{2}+C $

- $ =\frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+C $
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<footer style = "font-size: 18px">Problem-10 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-11 </span> $\rightarrow$ Problem-12 $\rightarrow$ Problem-13 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-13 </primary>
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- $ \int \frac{d x}{\sqrt{x^2+2 x+2}} $

- Solution:
- $ =\int \frac{d x}{\sqrt{(x+1)^2+1}} $

- [x+1=t]

- [d x=dt]

- $ =\int \frac{d t}{\sqrt{t^2+1}} $

- $ =\log \left|t+\sqrt{t^2+1}\right|+C $

- $ [ \because \int \frac{d x}{\sqrt{x^2+a^2}}=\log \mid x+ \sqrt{(x^2 + a^2)}+C ] $

- $ =\log \left|x+1+\sqrt{(x+1)^2+1}\right|+c $

- $ =\log \left|x+1+\sqrt{x^2+2 x+2}\right|+c $
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<footer style = "font-size: 18px">Problem-11 $\rightarrow$ Problem-12 $\rightarrow$ <span style = "color:blue"> Problem-13 </span> $\rightarrow$ Problem-14 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Problem-14 </primary>
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- $ I_1 =\int \frac{p x+q}{a x^2+b x+c} d x $

- $ I_2 =\int \frac{p x+q}{\sqrt{a x^2+b x+c}} d x $

- Solution:
- $ p x+q=A \frac{d}{d x}\left(a x^2+b x+c\right)+B $

- $ p x+q=A(2 a x+b)+B $

- $ p=2 a A, q=A b+B $

- $ A=? \quad B=? $

- $ I_1 =A \int \frac{\frac{d}{d x}\left(a x^2+b x+c\right)}{a x^2+b x+c} d x+B \int \frac{1}{a x^2+b x+c}dx $

- $ \rightarrow \text { can be evaluated. } $
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<footer style = "font-size: 18px">Problem-12 $\rightarrow$ Problem-13 $\rightarrow$ <span style = "color:blue"> Problem-14 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Indefinite Integral L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- [Put $3 x^2+2 x-1 = t]$   
- $ =\int \frac{d t}{t}-4 \times \frac{1}{3} \int \frac{d x}{x^2+\frac{2}{3} x-1} $

- $ [ x^2+\frac{2}{3} x-1 =\left(x+\frac{1}{3}\right)^2-\frac{4}{9} ] $

- $ =\log t+C_1-\frac{4}{3} \int \frac{d x}{(x+1 / 3)^2-(2 / 3)^2} $

- $ =\log \left|3 x^2+2 x-1\right|-\frac{4}{3} \cdot  \frac{1}{2 . \frac{2}{3}}. \log \left|\frac{x+1 / 3-2 / 3}{x+1 / 3+2 / 3}\right|+C_1+C_2 $

- $ =\log \left|3 x^2+2 x+1\right|-\log \left|\frac{3 x-1}{3 x+3}\right|+C $
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<footer style = "font-size: 18px">Problem-14 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thankyou $\rightarrow$  </footer>