Question

The integral $\int \frac{\sec ^2 x}{(\sec x+	an x)^{9 / 2}} d x$ equals (for some arbitrary constant $K$)

Accepted Answer

$-\frac{1}{(\sec x+	an x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+	an x)^2ight\}+K$

Answer

$-\frac{1}{(\sec x+	an x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+	an x)^2ight\}+K$

Answer

$\frac{1}{(\sec x+	an x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+	an x)^2ight\}+K$

Answer

$\frac{1}{(\sec x+	an x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+	an x)^2ight\}+K$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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