Maths Integrals

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$\begin{array}{l}7–6x–x^2=7–\left(x^2+6x+9–9\right)\end{array}$

$\begin{array}{l}=7–\left(x^2+6x+9\right)+9\end{array}$

$\begin{array}{l}=16–\left(x^2+6x+9\right) \end{array}$

$\begin{array}{l}=16–{\left(x+3\right)}^2\end{array}$

$\begin{array}{l}=\left(4^2\right)–{\left(x+3\right)}^2\end{array}$

$\begin{array}{l}I=\int \frac{1}{9x^2+6x+5}dx=\int \frac{1}{{(3x+1)}^2+2^2}\\ \begin{array}{l}Let \left(3x+1\right) =t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow 3dx=dt\hfill \end{array}\\ I=\int \frac{1}{{(3x+1)}^2+2^2}dx\\ =\frac{1}{3}\int \frac{1}{t^2+2^2}dt\\ =\frac{1}{3}\left[\frac{1}{2}ta{n}^{-1}\left(\frac{t}{2}\right)\right]+C\\ =\frac{1}{6}ta{n}^{-1}\left(\frac{t}{2}\right)+C=\frac{1}{6}ta{n}^{-1}\left(\frac{3x+1)}{2}\right)+C\end{array}$

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Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}\begin{array}{l}Let{x}^3=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}3x^{2}dx=dt\hfill \end{array}\\ I=\int \frac{x^2}{1-x^6}dx=\frac{1}{3}\int \frac{dt}{1-t^2}\\ =\frac{1}{3}\left[\frac{1}{2}log\left|\frac{1+t}{1-t}\right|\right]+C\\ =\frac{1}{6}log\left|\frac{1+x^3}{1-x^3}\right|+C\end{array}$

Kindly go through the solution

Kindly go through the solution