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The Integral [tex]\int \sec ^2\left(t\right)i+t\left(t^2+1\right)^4j+t^6\ln \left(t\right)kdt = i\tan \left(t\right)+\frac{j\left(t^2+1\right)^5}{10}+k\left(\frac{1}{7}t^7\ln \left(t\right)-\frac{t^7}{49}\right)[/tex]
Given integral;
[tex]\int \sec ^2\left(t\right)i+t\left(t^2+1\right)^4j+t^6\ln \left(t\right)kdt[/tex]
By solving the above integral, we have;
⇒ [tex]\int \sec ^2\left(t\right)i+t\left(t^2+1\right)^4j+t^6\ln \left(t\right)kdt[/tex]
By applying the sum rule,
⇒ [tex]\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx[/tex]
[tex]=\int \sec ^2\left(t\right)idt+\int \:t\left(t^2+1\right)^4jdt+\int \:t^6\ln \left(t\right)kdt[/tex]
[tex]\int \sec ^2\left(t\right)idt=i\tan \left(t\right)[/tex]
[tex]\int \:t\left(t^2+1\right)^4jdt = \frac{j\left(t^2+1\right)^5}{10}[/tex]
[tex]\int \:t^6\ln \left(t\right)kdt = k\left(\frac{1}{7}t^7\ln \left(t\right)-\frac{t^7}{49}\right)[/tex]
⇒ [tex]i\tan \left(t\right)+\frac{j\left(t^2+1\right)^5}{10}+k\left(\frac{1}{7}t^7\ln \left(t\right)-\frac{t^7}{49}\right)[/tex]
Therefore, the integral [tex]\int \sec ^2\left(t\right)i+t\left(t^2+1\right)^4j+t^6\ln \left(t\right)kdt = i\tan \left(t\right)+\frac{j\left(t^2+1\right)^5}{10}+k\left(\frac{1}{7}t^7\ln \left(t\right)-\frac{t^7}{49}\right)[/tex]
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