Question

$\mathfrak{PYQ EAMCET 2016}$
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Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{\int\,log\left(a^2+x^2\right)\,dx=h(x)+c}$

$\displaystyle\tt{\implies\,log\left(a^2+x^2\right)\int\,dx-\int\left[\dfrac{d}{dx}\left\{log\left(a^2+x^2\right)\right\}\cdot\int\,dx\right]dx=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-\int\left[\dfrac{2x}{a^2+x^2}\cdot\,x\right]dx=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-\int\dfrac{2x^2}{a^2+x^2}\,dx=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-\int\dfrac{2x^2+2a^2-2a^2}{a^2+x^2}\,dx=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-\int\dfrac{2x^2+2a^2}{a^2+x^2}\,dx-\int\dfrac{-2a^2}{a^2+x^2}\,dx=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-2\int\dfrac{a^2+x^2}{a^2+x^2}\,dx+2a^2\int\dfrac{dx}{a^2+x^2}=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-2\int\,dx+2a^2\int\dfrac{dx}{a^2+x^2}=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-2x+2a^2\cdot\dfrac{1}{a}\,tan^{-1}\left(\dfrac{x}{a}\right)+c=h(x)+c}$

$\displaystyle\tt{\implies\,x\cdot\,log\left(a^2+x^2\right)-2x+2a\,tan^{-1}\left(\dfrac{x}{a}\right)+c=h(x)+c}$

On comparing

$\displaystyle\tt{h(x)=x\cdot\,log\left(a^2+x^2\right)-2x+2a\,tan^{-1}\left(\dfrac{x}{a}\right)}$