Mathematics

# $\int {\dfrac{{2x + 7}}{{{{(x - 4)}^2}}}dx}$

##### SOLUTION

We have,

$I=\int{\dfrac{2x+7}{{{\left( x-4 \right)}^{2}}}dx}$

Let $t=x-4$

$dt=dx$

Therefore,

$I=\int{\dfrac{2\left( t+4 \right)+7}{{{t}^{2}}}dt}$

$I=\int{\dfrac{2t+8+7}{{{t}^{2}}}dt}$

$I=\int{\dfrac{2t+15}{{{t}^{2}}}dt}$

$I=\int{\dfrac{2t}{{{t}^{2}}}dt}+\int{\dfrac{15}{{{t}^{2}}}dt}$

$I=\ln \left( {{t}^{2}} \right)-\dfrac{15}{t}+C$

On putting the value of $t$, we get

$I=\ln \left( {{\left( x-4 \right)}^{2}} \right)-\dfrac{15}{\left( x-4 \right)}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 124

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