-x2 d x-a-b x2-A- 1-b2-x-2 a-blog a-b x-a2-b1-a-b x-B- 1-b2-x-2 a-blog a-b x-a2-b1-a-b x-C- 1-b2-x-2 a-blog a-b x-a2-b1-a-b x-D- 1-b2-x-a-b-2 a-blog a-b x-a2-b1-a-b x-

The correct option is D 1/b^2 [ x+a/b 2a/blog(a+bx) a^2/b1/(a+bx) ]Put a+bx=t ⇒ x = t a/b and dx = dt/b ∴ I=∫ ( t a/b )^2 ×1/t^2dt/b =1/b^3∫ ( 1 2a/t+a^2.t^ 2 ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

∫x2 d x/a+b x...

Question

$\int \frac{x^{2} d x}{(a + b x)^{2}} =$

\frac{1}{b^{2}} [x + \frac{2 a}{b} l o g (a + b x) - \frac{a^{2}}{b} \frac{1}{(a + b x)}]

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\frac{1}{b^{2}} [x - \frac{2 a}{b} l o g (a + b x) + \frac{a^{2}}{b} \frac{1}{(a + b x)}]

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\frac{1}{b^{2}} [x + \frac{2 a}{b} l o g (a + b x) + \frac{a^{2}}{b} \frac{1}{(a + b x)}]

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\frac{1}{b^{2}} [x + \frac{a}{b} - \frac{2 a}{b} l o g (a + b x) - \frac{a^{2}}{b} \frac{1}{(a + b x)}]

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Solution

The correct option is D $\frac{1}{b^{2}} [x + \frac{a}{b} - \frac{2 a}{b} l o g (a + b x) - \frac{a^{2}}{b} \frac{1}{(a + b x)}]$
Put $a + b x = t \Rightarrow x = \frac{t - a}{b}$ and $d x = \frac{d t}{b}$
$∴ I = \int {(\frac{t - a}{b})}^{2} \times \frac{1}{t^{2}} \frac{d t}{b}$
$= \frac{1}{b^{3}} \int (1 - \frac{2 a}{t} + a^{2} . t^{- 2}) d t = \frac{1}{b^{3}} [t - 2 a l o g t - \frac{a^{2}}{t}]$
$= \frac{1}{b^{2}} [x + \frac{a}{b} - \frac{2 a}{b} l o g (a + b x) - \frac{a^{2}}{b} \frac{1}{(a + b x)}]$

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∫x2 dx(a+bx)2=

The correct option is D 1b2[x+ab−2ablog(a+bx)−a2b1(a+bx)]Put a+bx=t⇒x=t−ab and dx=dtb ∴I=∫(t−ab)2×1t2dtb =1b3∫(1−2at+a2.t−2)dt=1b3[t−2a log t−a2t] =1b2[x+ab−2ablog(a+bx)−a2b1(a+bx)]

$\int \frac{x^{2} d x}{(a + b x)^{2}} =$