Question

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

Explanation:

(13)

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

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 \( \int t^{n}dt=\frac{t^{n + 1}}{n+1}+C\) (\(n
eq - 1\))
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:

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

(14)