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Question Number 124046 by john_santu last updated on 30/Nov/20

$$\varphi \left(\alpha \right)=\int \frac{6{\alpha }^2+30\alpha +2}{4{\alpha }^2+20\alpha +25}d\alpha$$

Answered by liberty last updated on 30/Nov/20

$$\varphi \left(\alpha \right)=\int \frac{6{\alpha }^2+30\alpha +2}{{(2\alpha +5)}^2}d\alpha$$$$\varphi \left(\alpha \right)=\int \frac{6\alpha (2\alpha +5)-2(3{\alpha }^2-1)}{{(2\alpha +5)}^2}d\alpha$$$$\varphi \left(\alpha \right)=\int \frac{d}{d\alpha }\left(\frac{3{\alpha }^2-1}{2\alpha +5}\right).d\alpha =\int d\left(\frac{3{\alpha }^2-1}{2\alpha +5}\right)$$$$\varphi \left(\alpha \right)=\frac{3{\alpha }^2-1}{2\alpha +5}+c.$$

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