int frac{log left(x^2+a^2right)}{x^2} ,d x= | vedclass

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Question

(B) माना $I = \int \frac{\log \left(x^2+a^2\right)}{x^2} \,d x$.खंडशः समाकलन (Integration by parts) का उपयोग करते हुए,$\int u v \,d x = u \int v \,d x

Vedclass · Accepted Answer

$-\frac{\log \left(x^2+a^2ight)}{x}+\frac{2}{a} 	an ^{-1} \frac{x}{a}+c$,जहाँ $c$ एक समाकलन स्थिरांक है।

Vedclass · Answer

(B) माना $I = \int \frac{\log \left(x^2+a^2ight)}{x^2} \,d x$.खंडशः समाकलन (Integration by parts) का उपयोग करते हुए,$\int u v \,d x = u \int v \,d x - \int \left( \frac{du}{dx} \int v \,d x ight) d x$.माना $u = \log \left(x^2+a^2ight)$ और $v = x^{-2}$.तब $\frac{du}{dx} = \frac{2x}{x^2+a^2}$ और $\int v \,d x = -\frac{1}{x}$.$I = \log \left(x^2+a^2ight) \cdot \left(-\frac{1}{x}ight) - \int \left( \frac{2x}{x^2+a^2} \cdot \left(-\frac{1}{x}ight) ight) d x$.$I = -\frac{\log \left(x^2+a^2ight)}{x} + 2 \int \frac{1}{x^2+a^2} \,d x$.मानक समाकलन $\int \frac{1}{x^2+a^2} \,d x = \frac{1}{a} 	an^{-1} \left(\frac{x}{a}ight) + c$ का उपयोग करते हुए.$I = -\frac{\log \left(x^2+a^2ight)}{x} + \frac{2}{a} 	an^{-1} \left(\frac{x}{a}ight) + c$.

$\int \frac{\log \left(x^2+a^2\right)}{x^2} \,d x=$

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