<p><strong>261. <math><mrow><mstyle displaystyle="true"><mrow><msubsup><mo>∫</mo><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></msubsup><mrow><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></mfrac></mrow></mrow></mstyle><mo>.</mo></mrow></math> <math><mrow><mo>=</mo><mfrac><mrow><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>+</mo><mi>l</mi><mi>o</mi><mi>g</mi><mfrac><mrow><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></math></strong></p>

Let I=∫13dxx2(x+1).The integrand is of the form.1 = Ax (x + 1) + B (x + 1) + Cx2= A (x2 + x) + B (x + 1) + Cx2Comparing the coefficients,A + C = 0 ____ (1)A + B = 0 ______ (2)B = 1 ________ (3)Putting Equation (3) in (2),A + 1 = 0A = -1.and putting value of A in Equation (1),-1 + C = 0C = 11x2(x+1)=−1x+1x2+1x+1∴I=∫13dxx2(x+1)=∫13−dxx+∫13dxx2+∫13dxx+1=− [log|x|]13+[x−2+1−2+1]13+[log?x+1]13=[−log3+log1]−[1x]13+[log|3+1|−log|1+1|]=−log3+0−[13−1]+log4−log2=−log3−(1−3)3+log22−log2=−log3−(−2)3+2log2−log2=log2−log3+23=23+log23Hence proved.

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261. $\int_{1}^{3} \frac{d x}{x^{2} (x + 1)} .$ $= \frac{2}{3} + l o g \frac{2}{3}$

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Vishal Baghel | Contributor-Level 10

4 months ago

Let $I = \int_{1}^{3} \frac{d x}{x^{2} (x + 1)} .$

The integrand is of the form.

1 = Ax (x + 1) + B (x + 1) + Cx²

= A (x² + x) + B (x + 1) + Cx²

Comparing the coefficients,

A + C = 0 ____ (1)

A + B = 0 ______ (2)

B = 1 ________ (3)

Putting Equation (3) in (2),

A + 1 = 0

A = -1.

and putting value of A in Equation (1),

-1 + C = 0

C = 1

$\frac{1}{x^{2} (x + 1)} = \frac{- 1}{x} + \frac{1}{x^{2}} + \frac{1}{x + 1}$

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

$= - {[l o g | x |]}_{1}^{3} + {[\frac{x^{- 2 + 1}}{- 2 + 1}]}_{1}^{3} + {[l o g ? x + 1]}_{1}^{3}$

$= [- l o g 3 + l o g 1] - {[\frac{1}{x}]}_{1}^{3} + [l o g | 3 + 1 | - l o g | 1 + 1 |]$

$= - l o g 3 + 0 - [\frac{1}{3} - 1] + l o g 4 - l o g 2$

$= - l o g 3 - \frac{(1 - 3)}{3} + l o g 2^{2} - l o g 2$

$= - l o g 3 - \frac{(- 2)}{3} + 2 l o g 2 - l o g 2$

$= l o g 2 - l o g 3 + \frac{2}{3}$

$= \frac{2}{3} + l o g \frac{2}{3}$

Hence proved.

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261. $\int_{1}^{3} \frac{d x}{x^{2} (x + 1)} .$ $= \frac{2}{3} + l o g \frac{2}{3}$

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261. $\int_{1}^{3} \frac{d x}{x^{2} (x + 1)} .$ $= \frac{2}{3} + l o g \frac{2}{3}$