Question

Solving integrals. Please explain how you got the answer. Thank you!
1. ∫e^(-2x)*(3+4e^(-2x))^6 dx
2. ∫ x/(9+16x^2) dx
3. ∫ (3x+1)/sqrt(5x^2+3) dx
4. ∫ x/sqrt(5+4x-x^2) dx

Answer 1

\[I=\int\limits e ^{-2x}*\left\{ 3+4e ^{-2x} \right\}^{6}dx\]
\[put~ 3+4e ^{-2x}=y,e ^{-2x}=\frac{ y-3 }{4 },e ^{-2x}\left( -2 \right)dx=\frac{ 1 }{ 4 }dy\]
\[e ^{-2x}dx=-\frac{ 1 }{ 8 }dy\]
\[I=-\frac{ 1 }{ 8 } \int\limits y^6 dy\]
complete it.

Answer 2

4. ∫ x/sqrt(5+4x-x^2) dx

\[(1)~~\int e^{-2x}\left(3+4e^{-2x}\right)^6~dx\]
Substitute \(u=3+4e^{-2x}\), then \(du=-8e^{-2x}~dx\), or \(-\dfrac{1}{8}du=e^{-2x}~dx\):
\[-\frac{1}{8}\int u^6~du=\cdots\]
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\[(2)~~\int\frac{x}{9+16x^2}~dx\]
Substitute \(t=9+16x^2\), then \(dt=32x~dx\), or \(\dfrac{1}{32}dt=x~dx\):
\[\frac{1}{32}\int\frac{dt}{t}~dt=\cdots\]
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\[(3)~~\int\frac{3x+1}{\sqrt{5x^2+3}}~dx=3\int\frac{x}{\sqrt{5x^2+3}}~dx+\int\frac{dx}{\sqrt{5x^2+3}}\]
For the first integral, substitute \(p=5x^2+3\), then \(dp=10x~dx\), or \(\dfrac{1}{10}dp=x~dx\). For the second integral, substitute \(x=\sqrt{\dfrac{3}{5}}\tan r\), then \(dx=\sqrt{\dfrac{3}{5}}\sec^2r~dr\).
\[\frac{3}{10}\int\frac{dp}{\sqrt{p}}+\int\frac{\sqrt{\frac{3}{5}}\sec^2r}{\sqrt{5\left(\sqrt{\frac{3}{5}}\tan r\right)^2+3}}~dr\\
\frac{3}{10}\int p^{-1/2}~dp +\sqrt{\frac{3}{5}}\int\frac{\sec^2r}{\sqrt{5\left(\frac{3}{5}\tan^2 r\right)+3}}~dr\\
\frac{3}{10}\int p^{-1/2}~dp +\sqrt{\frac{3}{5}}\int\frac{\sec^2r}{\sqrt3\sqrt{\tan^2 r+1}}~dr\\
\frac{3}{10}\int p^{-1/2}~dp +\frac{1}{\sqrt{5}}\int\frac{\sec^2r}{\sqrt{\sec^2 r}}~dr\\
\frac{3}{10}\int p^{-1/2}~dp +\frac{1}{\sqrt{5}}\int \sec r~dr=\cdots\]
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\[(4)~~\int\frac{x}{\sqrt{5+4x-x^2}}~dx\]
Complete the square in the denominator:
\[\begin{align*}5+4x-x^2&=-\left(x^2-4x-5\right)\\
&=-\left(x^2-4x+4-9\right)\\
&=-\left((x-2)^2-9\right)\\
&=9-(x-2)^2\end{align*}\]
\[\int\frac{x}{\sqrt{9-(x-2)^2}}~dx\]
Substitute \(w=x-2\), then \(dw=dx\):
\[\int\frac{w+2}{\sqrt{9-w^2}}~dw=\int\frac{w}{\sqrt{9-w^2}}~dw+2\int\frac{dw}{\sqrt{9-w^2}}\]
For the first integral, substitute \(q=9-w^2\), then \(dq=-2w~dw\), or \(-\dfrac{1}{2}dq=w~dw\). For the second integral, substitute \(w=3\sin z\), then \(dw=3\cos z~dz\):
\[-\frac{1}{2}\int\frac{dq}{\sqrt{q}}+2\int\frac{3\cos z}{\sqrt{9-(3\sin z)^2}}~dz\\
-\frac{1}{2}\int q^{-1/2}~dq+\frac{6}{\sqrt9}\int\frac{\cos z}{\sqrt{1-\sin^2z}}~dz\\
-\frac{1}{2}\int q^{-1/2}~dq+\frac{6}{\sqrt9}\int\frac{\cos z}{\sqrt{\cos^2z}}~dz\\
-\frac{1}{2}\int q^{-1/2}~dq+\frac{6}{\sqrt9}\int dz=\cdots\]