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Class: subject: hence evaluate (i) or (ii)) \[ \begin{array}{l} \therefore I=\sqrt{x^{2}+a^{2}} \int 1 d x-\int\left[\frac{d\left(\sqrt{x^{2}+a^{2}}\right)}{d x} \cdot \int 1 d x\right] d x \\ =x \sqrt{x^{2}+a^{2}}-\int\left[\frac{1}{2 \sqrt{x^{2}+a^{2}}}(2 x) \cdot x\right] d x \\ =x \sqrt{x^{2}+a^{2}}-\int \frac{\left(x^{2}+a^{2}-a^{2}\right.}{\sqrt{x^{2}+a^{2}}} d x . \\ \text { – }=x \sqrt{x^{2}+a^{2}}-\int \frac{x^{2}+a^{2}}{\sqrt{x^{2}+a^{2}}} d x+a^{2} \int \frac{d x}{\sqrt{x^{2}+a^{2}}} \\ \therefore I=x \sqrt{x^{2}+a^{2}}-\int \sqrt{x^{2}+a^{2}} d x+a^{2} \int \frac{d x}{\sqrt{x^{2}+a^{2}}} \text {. } \\ {\left[\because \int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c .\right]} \\ \end{array} \]
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| Question Text |
Class: subject: hence evaluate (i) or (ii)) \[ \begin{array}{l} \therefore I=\sqrt{x^{2}+a^{2}} \int 1 d x-\int\left[\frac{d\left(\sqrt{x^{2}+a^{2}}\right)}{d x} \cdot \int 1 d x\right] d x \\ =x \sqrt{x^{2}+a^{2}}-\int\left[\frac{1}{2 \sqrt{x^{2}+a^{2}}}(2 x) \cdot x\right] d x \\ =x \sqrt{x^{2}+a^{2}}-\int \frac{\left(x^{2}+a^{2}-a^{2}\right.}{\sqrt{x^{2}+a^{2}}} d x . \\ \text { – }=x \sqrt{x^{2}+a^{2}}-\int \frac{x^{2}+a^{2}}{\sqrt{x^{2}+a^{2}}} d x+a^{2} \int \frac{d x}{\sqrt{x^{2}+a^{2}}} \\ \therefore I=x \sqrt{x^{2}+a^{2}}-\int \sqrt{x^{2}+a^{2}} d x+a^{2} \int \frac{d x}{\sqrt{x^{2}+a^{2}}} \text {. } \\ {\left[\because \int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c .\right]} \\ \end{array} \] |
| Updated On | Jan 13, 2023 |
| Topic | Calculus |
| Subject | Mathematics |
| Class | Class 12 Passed |
| Answer Type | Video solution: 1 |
| Upvotes | 113 |
| Avg. Video Duration | 8 min |