(13) $\\int\\frac{7x^{6}+3}{3x + x^{7}}dx$\n(14) $\\int\\frac{\\cos x}{\\sin^{4}x}dx$\n

(13)

Explanation:

Step1: Let ( u = 3x + x^{7} )

Differentiate ( u ) with respect to ( x ): ( du=(3 + 7x^{6})dx )

Step2: Substitute into the integral

The integral ( \int\frac{7x^{6}+3}{3x + x^{7}}dx=\int\frac{du}{u} )

Step3: Integrate

Using the formula ( \int\frac{1}{u}du=\ln|u|+C ) (where ( C ) is the constant of integration) Since ( u = 3x + x^{7} ), the integral becomes ( \ln|3x + x^{7}|+C )

Answer:

( \ln|x^{7}+3x| + C )

(14)

Explanation:

Step1: Let ( t=\sin x )

Differentiate ( t ) with respect to ( x ): ( dt=\cos xdx )

Step2: Substitute into the integral

The integral ( \int\frac{\cos x}{\sin^{4}x}dx=\int\frac{dt}{t^{4}}=\int t^{- 4}dt )

Step3: Integrate

Using the formula ( \int t^{n}dt=\frac{t^{n + 1}}{n+1}+C) ((n\neq - 1)) Here (n=-4), so ( \int t^{-4}dt=\frac{t^{-4 + 1}}{-4+1}+C=-\frac{1}{3t^{3}}+C )

Step4: Back - substitute

Since ( t = \sin x ), the integral is ( -\frac{1}{3\sin^{3}x}+C=-\frac{1}{3}\csc^{3}x + C)

Answer:

( -\frac{1}{3}\csc^{3}x + C )