100 Integral Problems and Solutions (part 1)

Hallo Guys
Assalamu'alaikum Wr.Wb.
Nama saya Hasbiansyah Cahyadi
Disini saya akan membahas soal² integral,, Mari simak pembahasannya
Berikut adalah 100 soal integral yang akan saya bahas
Pembahasan ini akan saya jadikan 4 bagian, untuk soal ini adalah problem no 1-25

Soal Integral Aljabar dan Trigonometri

Ini adalah pembahasan part ke 1, yaitu soal no 1-25

1) $\int\frac{1}{\sqrt{x}(x+1)}\ dx$
Jawab : Misalkan $u=\sqrt{x}$
                              $du=\frac {1}{2\sqrt{x}}\ dx$
$\int\frac{1}{\sqrt{x}(x+1)}\ dx$
=$\int\frac{2}{x+1}.\frac{1}{2\sqrt{x}}\ dx$
=$\int\frac{2}{u^2+1}\ du$
Substitusi $u=tan\ y$ maka $du=sec^2\ y \ dy$ sehingga
=$\int\frac{2}{tan^2\ y+1}.sec^2\ y\ dy$
=$\int\frac{2}{sec^2\ y}.sec^2\ y\ dy$
=$\int 2\ dy$
=$2y +C=2tan^{-1}\ u +C$
=$2tan^{-1}\left(\sqrt{x}\right) +C$
2) $\int\frac{sec^2\ t}{1+tan \ t}\ dt$
Jawab : Misalkan $u=1+tan\ t$
                              $du=sec^2\ t\ dt$
$\int\frac{sec^2\ t}{1+tan\ t}\ dt$
=$\int\frac{1}{u}\ du$
=$ln |u| +C$
=$ln |1+tan\ t| +C$
3) $\int sin\ x.sec\ x\ dx$
Jawab : Kita tau bahwa $sec\ x=\frac{1}{cos\ x}$
$\int sin\ x.sec\ x dx$
=$\int\frac{sin\ x}{cos\ x}\ dx$
Misalkan $u=cos\ x$
                $du=-sin\ x\ dx$
=$\int\frac{-1}{u}\ du$
=$-ln|u| +C$
=$-ln|cos\ x| +C$
4) $\int\frac{csc\ x.cot\ x}{1+csc^2\ x}\ dx$
Jawab : Misalkan : $u=csc\ x$
                                $du=-csc\ x.cot\ x\ dx$
$\int\frac{csc\ x.cot\ x}{1+csc^2\ x}\ dx$
=$\int\frac{-1}{1+u^2}\ du$
sesuai dengan langkah no 1, diperoleh
=$-tan^{-1}\ u +C$
=$-tan^{-1}\ (csc\ x) +C$
5) $\int\frac{tan\ \theta}{cos^2\ \theta}\ d\theta$
Jawab : Karena $tan\ \theta=\frac{sin\ \theta}{cos\ \theta}$ maka
$\int\frac{tan\ \theta}{cos^2\ \theta}\ d\theta=\int\frac{sin\ \theta}{cos^3\ \theta}\ d\theta$
Misalkan $u=cos\ \theta$
                $du=-sin\ \theta\ d\theta$
sehingga bisa kita tulis
=$\int\frac{-1}{u^3}\ du$
=$\frac{1}{2u^2} +C$
=$\frac{1}{2cos^2\ \theta} +C$
6) $\int csc^4\ x\ dx$
Jawab : Soal ekuivalen dengan $\int csc^2\ x(1+cot^2\ x)\ dx$
Dimisalkan $u=cot\ x$
                    $du=-csc^2\ x\ dx$
$\int csc^2\ x(1+cot^2\ x)\ dx$
=$\int (1+u^2)\ -du$
=$-u-\frac{u^3}{3} +C$
=$-cot\ x-\frac{cot^3\ x}{3} +C$
7) $\int x.tan^2\ x\ dx$
Jawab : Misalkan $\begin{align*}u=x\\                               du&=dx\\                   dv&=tan^2\ x\ dx\\                 v&=\int tan^2\ x\ dx\\                   &=\int (sec^2\ x -1)\ dx\\                   &=tan\ x -x\end{align*}$

Sehingga
$\int x.tan^2\ x\ dx=\int u\ dv=uv-\int v\ du$
=$x(tan\ x -x)-\int tan\ x -x\ dx$
Berdasarkan soal nomor 3 kita nisa tau bahwa $\int tan\ x dx=ln |\frac{1}{cos\ x}|$
=$xtan\ x-x^2-(ln |\frac{1}{cos\ x}|-\frac{1}{2}x^2) +C$
=$xtan\ x-x^2+ln |cos\ x|+\frac{1}{2}x^2 +C$
=$xtan\ x+ln |cos\ x|-\frac{1}{2}x^2 +C$
8) $\int x^2.cos^2\ x\ dx$
Jawab : Kita tahu bahwa $cos\ 2x=2cos^2\ x-1$ atau $cos^2\ x=\frac{cos\ 2x+1}{2}$
$\int x^2.cos^2\ x\ dx$
=$\int x^2(\frac{cos\ 2x+1}{2})\ dx$
=$\int x^2.\frac{cos\ 2x}{2}\ dx +\int\frac{x^2}{2}\ dx$
Kita tahu hasil dari $\int\frac{x^2}{2}\ dx=\frac{x^3}{3}$
Obs: $\int x^2.\frac{cos\ 2x}{2}\ dx$
misalkan $u=\frac{x^2}{2}$ dan $dv=cos\ 2x\ dx$ maka $du=x\ dx$ dan $v=\frac{sin\ 2x}{2}$
$\int x^2.\frac{cos\ 2x}{2}\ dx=\int u\ dv$
=$uv-\int v\ du$
=$\frac{x^2}{2}.\frac{sin\ 2x}{2}-\int \frac{sin\ 2x}{2}.x\ dx$
dengan cara yg sama kita misalkan lagi $u=\frac{x}{2}$ dan $dv=sin\ 2x\ dx$ maka $du=\frac{1}{2}\ dx$ dan $v=-\frac{cos\ 2x}{2}$
$\frac{x^2}{2}.\frac{sin\ 2x}{2}-\int \frac{sin\ 2x}{2}.x\ dx=\frac{x^2}{2}.\frac{sin\ 2x}{2}-\int u\ dv$
=$\frac{x^2}{2}.\frac{sin\ 2x}{2}-uv+\int v\ du$
=$\frac{x^2}{2}.\frac{sin\ 2x}{2}-\frac{x}{2}.(-\frac{cos\ 2x}{2})+\int -\frac{cos\ 2x}{2}.\frac{1}{2}\ dx$
=$\frac{x^2.sin\ 2x}{4}+\frac{x.cos\ 2x}{4}-\frac{1}{4}\int cos\ 2x\ dx$
=$\frac{x^2.sin\ 2x}{4}+\frac{x.cos\ 2x}{4}-\frac{1}{8}.sin\ 2x +C$
Sehingga nilai dari
 $\int x^2.cos^2\ x\ dx=\frac{x^3}{3}+\frac{x^2.sin\ 2x}{4}+\frac{x.cos\ 2x}{4}-\frac{1}{8}.sin\ 2x +C$
9) $\int x^5\sqrt{2-x^3}\ dx$
Jawab : Misalkan $u=2-x^3$
                              $du=-\frac{x^2}{3}\ dx$
Perhatikan bahwa $\int x^5\sqrt{2-x^3}\ dx=\int -3x^3\sqrt{2-x^3}.(-\frac{x^2}{3})\ dx=\int -3(2-u)\sqrt{u}\ du$
=$\int -6\sqrt{u}+\int 3u^{\frac{3}{2}}\ du$
=$-4u^{\frac{3}{2}}+\frac{6}{5}u^{\frac{5}{2}} +C$
=$-4(2-x^3)^{\frac{3}{2}}+\frac{6}{5}(2-x^3)^{\frac{5}{2}} +C$
10) $\int\frac{1}{x^2+4}\ dx$
Jawab : Lakukan substitusi $x=2tan\ u$
                                                $dx=2sec^2\ u$
$\int\frac{1}{x^2+4}\ dx$
=$\int\frac{1}{(2tan\ u)^2+4}.2sec^2\ u\ du$
=$\int\frac{1}{4tan^2\ u+4}.2sec^2\ u\ du$
=$\int\frac{1}{4(tan^2\ u+1)}.2sec^2\ u\ du$
=$\int\frac{1}{4sec^2\ u}.2sec^2\ u\ du$
=$\int\frac{1}{2}\ du$
=$\frac{u}{2} +C$
=$\frac{tan^{-1}\frac{x}{2}}{2} +C$
11) $\int\frac{x^2}{\sqrt{25+x^2}}\ dx$
Jawab : Lakukan substitusi $\begin{align*}x&=5tan\ u\\                         dx&=5sec^2\ u\ du\end{align*}$
$\int\frac{x^2}{\sqrt{25+x^2}}\ dx$
=$\int\frac{(5tan\ u)^2.5sec^2\ u}{\sqrt{25+(5tan\ u)^2}}\ du$
=$\int\frac{125tan^2\ u.sec^2\ u}{\sqrt{25(1+tan^2\ u)}}\ du$
=$\int\frac{125tan^2\ u.sec^2\ u}{\sqrt{25sec^2\ u}}\ du$
=$\int\frac{125tan^2\ u.sec^2\ u}{5sec\ u}\ du$
=$\int 25(sec^2\ u-1).sec\ u\ du$
=$25\int sec^3\ u\ du -25\int sec\ u\ du$
Obs: $\int sec^3\ u\ du$
=$\int sec\ u.sec^2\ u\ du$
misal : $p=sec\ u$ dan $dq=sec^2\ u\ du$
           $dp=sec\ u.tan\ u\ du$ dan $q=tan\ u$
$\int sec\ u.sec^2\ u\ du=m$
$m=\int p\ dq$
$m=pq-\int q\ dp$
$m=sec\ u.tan\ u-\int tan^2\ u.sec\ u\ du$
$m=sec\ u.tan\ u-\int (sec^2\ u-1)sec\ u du$
$m=sec\ u.tan\ u-\int sec^3\ u\ du+\int sec\ u du$
$m=sec\ u.tan\ u-m+\int sec\ u du$
$m=\frac{sec\ u.tan\ u+\int sec\ u du}{2}$
$\int sec^3\ u\ du=\frac{sec\ u.tan\ u+ln|tan\ u+sec\ u|}{2}$
Obs: $\int sec\ u\ du=ln|tan\ u+sec\ u|$
Jadi,
$25\int sec^3\ u\ du -25\int sec\ u\ du$
=$25\left(\frac{sec\ u.tan\ u+ln|tan\ u+sec\ u|}{2})-25(ln|tan\ u+sec\ u|\right)+C$
=$25\left(\frac{sec\ u.tan\ u-ln|tan\ u+sec\ u|}{2}\right)+C$
=$\frac{25}{2}\left(\frac{\sqrt{x^2+25}}{5}.\frac{x}{5}-ln|\frac{x}{5}+\frac{\sqrt{x^2+25}}{5}|\right)+C$
=$\frac{\left(x\sqrt{x^2+25}-25ln|x+\sqrt{x^2+25}|\right)}{2}+C$
12) $\int (cos\ x)\sqrt{4-sin^2\ x}\ dx$
Jawab :
$\int (cos\ x)\sqrt{4-sin^2\ x}\ dx$
=$\int (cos\ x)\sqrt{(2+sin\ x)(2-sin\ x)}\ dx$
misalkan : $u=2+sin\ x$ maka $2-sin\ x=4-u$
                  $du=cos\ x\ dx$
=$\int (cos\ x)\sqrt{(2+sin\ x)(2-sin\ x)}\ dx$
=$\int \sqrt{u(4-u)}\ du$
=$\int \sqrt{4-(u-2)^2}\ du$
misalkan : $u-2=2sin\ y$
                  $du=2cos\ y\ dy$
$\int \sqrt{4-(u-2)^2}\ du$
=$\int \sqrt{4-4sin^2\ y}.2cos\ y\ dy$
=$\int \sqrt{4cos^2\ y}.2cos\ y\ dy$
=$\int 4cos^2\ y\ dy$
=$\int 2(cos\ 2y+1) dy$
=$sin\ 2y+2y +C$
=$\frac{u-2}{2}+2sin^{-1}\left(\frac{u-2}{2}\right) +C$
=$\frac{sin\ x}{2}+2sin^{-1}\left(\frac{sin\ x}{2}\right) +C$
13) $\int\frac{1}{x^2-x+1}\ dx$
Jawab :
$\int\frac{1}{x^2-x+1}\ dx$
=$\int\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}\ dx$
Misalkan : $x-\frac{1}{2}=\frac{\sqrt{3}}{2}u$
                  $dx=\frac{\sqrt{3}}{2}\ du$
$\int\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}\ dx$
=$\int\frac{1}{\left(\frac{\sqrt{3}}{2}u\right)^2+\frac{3}{4}}.\frac{\sqrt{3}}{2}\ du$
=$\int\frac{4}{3(u^2+1)}.\frac{\sqrt{3}}{2}\ du$
=$\frac{2\sqrt{3}}{3}\int\frac{1}{u^2+1}\ du$
=$\frac{2\sqrt{3}}{3}tan^{-1}\ u+C$
=$\frac{2\sqrt{3}}{3}tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$
14) $\int \sqrt{x^2+x+1}\ dx$
Jawab :
$\int \sqrt{x^2+x+1}\ dx$
=$\int \sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}\ dx$
Misalkan : $x+\frac{1}{2}=\frac{\sqrt{3}}{2}u$
                  $dx=\frac{\sqrt{3}}{2}\ du$
=$\int \sqrt{(\frac{\sqrt{3}}{2}u)^2+\frac{3}{4}}.\frac{\sqrt{3}}{2}\ du$
=$\int\frac{3}{4}\sqrt{u^2+1}\ du$
Substitusi $u=tan\ y$
                 $du=sec^2\ y\ dy$
=$\int\frac{3}{4}\sqrt{tan^2\ y+1}.sec^2\ y\ dy$
=$\int\frac{3}{4}sec^3\ y\ dy$
Disoal no 11 sudah saya hitung nilainya
=$\frac{3}{4}\frac{sec\ y.tan\ y+ln|tan\ y+sec\ y|}{2}+ C$
=$\frac{3}{4}\frac{\sqrt{1+u^2}.u+ln\left|\sqrt{1+u^2}+u\right|}{2}+ C$
=$\frac{3}{4}\frac{\sqrt{1+\left(\frac{2x+1}{\sqrt{3}}\right)^2}.\frac{2x+1}{\sqrt{3}}+ln\left|\sqrt{1+\left(\frac{2x+1}{\sqrt{3}}\right)^2}+\frac{2x+1}{\sqrt{3}}\right|}{2}+ C$
15) $\int\frac{5x+31}{3x^2-4x+11}\ dx$
Jawab : Perhatikan bahwa
$\int\frac{5x+31}{3x^2-4x+11}\ dx$
=$\int\frac{5x-\frac{10}{3}+\frac{10}{3}+31}{3x^2-4x+11}\ dx$
=$\int\frac{5x-\frac{10}{3}}{3x^2-4x+11}\ dx+\frac{103}{3}\int\frac{1}{3x^2-4x+11}\ dx$
Obs : $\int\frac{5x-\frac{10}{3}}{3x^2-4x+11}\ dx$
misalkan : $\begin{align*}u&=3x^2-4x+11\\      \frac{5}{6}du &=\frac{5}{6}(6x-4)\ dx\\                &=5x-\frac{10}{3}\ dx\end{align*}$
$\int\frac{5x-\frac{10}{3}}{3x^2-4x+11}\ dx$
=$\int\frac{5}{6u}\ du$
=$\frac{5}{6}ln|u|$
=$\frac{5}{6}ln|3x^2-4x+11|$
Obs : $\frac{103}{3}\int\frac{1}{3x^2-4x+11}\ dx$
=$103\int\frac{1}{9x^2-12x+33}\ dx$
=$103\int\frac{1}{(3x-2)^2+29}\ dx$
misalkan $u=\frac{3x-2}{\sqrt{29}}$
                $du=\frac{3}{\sqrt{29}}\ dx$
                $dx=\frac{\sqrt{29}}{3}\ du$
=$103\int\frac{\sqrt{29}}{3(29u^2+29)}\ du$
=$\frac{103\sqrt{29}}{87}\int\frac{1}{u^2+1}\ du$
=$\frac{103\sqrt{29}}{87}tan^{-1}\ u$
=$\frac{103\sqrt{29}}{87}tan^{-1}\left(\frac{3x-2}{\sqrt{29}}\right)$
Jadi,
$\int\frac{5x+31}{3x^2-4x+11}\ dx$
=$\int\frac{5x-\frac{10}{3}}{3x^2-4x+11}\ dx+\frac{103}{3}\int\frac{1}{3x^2-4x+11}\ dx$
=$\frac{5}{6}ln|3x^2-4x+11|+\frac{103\sqrt{29}}{87}tan^{-1}\left(\frac{3x-2}{\sqrt{29}}\right)+C$
16) $\int\frac{x^4+1}{x^2+2}\ dx$
Jawab : Perhatikan bahwa
$x^4+1=\left(x^2+2\right)^2-4x^2-3              =\left(x^2+2\right)^2-4\left(x^2+2\right)+5$
$\int\frac{x^4+1}{x^2+2}\ dx$
=$\int (x^2+2)-4+\frac{5}{x^2+2}\ dx$
=$\int x^2-2+\frac{5}{x^2+2}\ dx$
=$\frac{x^3}{3}-2x+5\int\frac{1}{x^2+2}\ dx$
Obs : $\int\frac{1}{x^2+2}\ dx$
=$\int\frac{1}{x^2+2}\ dx$
misalkan $u=\frac{x}{\sqrt{2}}$
                $du=\frac{1}{\sqrt{2}}$
=$\int\frac{1}{\sqrt{2}\left(2u^2+2\right)}\ du$
=$\frac{1}{2\sqrt{2}}tan^{-1}\ u$
=$\frac{1}{2\sqrt{2}}tan^{-1}\left(\frac{x}{\sqrt{2}}\right)$
Jadi,
$\int\frac{x^4+1}{x^2+2}\ dx$
=$\frac{x^3}{3}-2x+5\int\frac{1}{x^2+2}\ dx$
=$\frac{x^3}{3}-2x+\frac{5}{2\sqrt{2}}tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+C$
17) $\int\frac{1}{5+4cos\ \theta}\ d\theta$
Jawab :
$\int\frac{1}{5+4cos\ \theta}\ d\theta$
=$\int\frac{1}{5+4(2cos^2\frac{\theta}{2}-1)}\ d\theta$
=$\int\frac{1}{1+8cos^2\frac{\theta}{2}}\ d\theta$

=$\int\frac{sec^2\frac{\theta}{2}}{sec^2\frac{\theta}{2}+8}\ d\theta$
=$\int\frac{sec^2\frac{\theta}{2}}{tan^2\frac{\theta}{2}+9}\ d\theta$
misalkan $u=\frac{tan\ \frac{\theta}{2}}{3}$
                $du=\frac{sec^2\ \frac{\theta}{2}}{6}\ d\theta$
=$\int\frac{6}{9u^2+9}\ du$
=$\frac{2}{3}\int\frac{1}{u^2+1}\ du$
=$\frac{2}{3}tan^{-1}\ u+C$
=$\frac{2}{3}tan^{-1}\left(\frac{tan\ \frac{\theta}{2}}{3}\right)+C$
18) $\int\frac{\sqrt{x}}{1+x}\ dx$
Jawab : misalkan $u=\sqrt{x}$
                              $du=\frac{1}{2\sqrt{x}}\ dx$
                              $dx=2\sqrt{x}\ du$
$\int\frac{\sqrt{x}}{1+x}\ dx$
=$\int\frac{2u^2}{1+u^2}\ du$
=$\int\frac{2(1+u^2)-2}{1+u^2}\ du$
=$\int\ 2-\frac{2}{1+u^2}\ du$
=$2u-2tan^{-1}\ u+C$
=$2\sqrt{x}-2tan^{-1}\ \sqrt{x}+C$
19) $\int\frac{cos\ x}{\sqrt{4-sin^2\ x}}\ dx$
Jawab :
$\int\frac{cos\ x}{\sqrt{4-sin^2\ x}}\ dx$
misalkan : $u=\frac{sin\ x}{2}$
                    $du=\frac{cos\ x}{2}\ dx$
=$\int\frac{2}{\sqrt{4-4u^2}}\ du$
=$\int\frac{1}{\sqrt{1-u^2}}\ du$
lakukan substitusi $u=sin\ y$
                                   $du=cos\ y\ dy$
=$\int\frac{cos\ y}{\sqrt{1-sin^2\ y}}\ dy$
=$\int 1 dy$
=$y+C$
=$sin^{-1}\ u+C$
=$sin^{-1}\left(\frac{sin\ x}{2}\right)+C$
20) $\int\frac{cos\ 2x}{cos\ x}\ dx$
Jawab :
$\int\frac{cos\ 2x}{cos\ x}\ dx$
=$\int\frac{2cos^2\ x-1}{cos\ x}\ dx$
=$\int 2cos\ x-sec\ x\ dx$
=$2sin\ x-ln|tan\ x +sec\ x|+C$
21) $\int\frac{tan\ x}{ln(cos\ x)}\ dx$
Jawab :
$\int\frac{tan\ x}{ln(cos\ x)}\ dx$
=$\int\frac{tan\ x}{-ln(sec\ x)}\ dx$
misalkan $u=ln(sec\ x)$
                  $du=tan\ x\ dx$
=$\int\frac{1}{-u}\ du$
=$-ln|u|+C$
=$-ln|ln(sec\ x)|+C$
22) $\int\frac{x^7}{\sqrt{1-x^4}}\ dx$
Jawab : misalkan $u=1-x^4$
                                $du=-4x^3\ dx$
$\int\frac{x^7}{\sqrt{1-x^4}}\ dx$
=$\int\frac{x^4}{-4\sqrt{u}}\ du$
=$\int\frac{1-u}{-4\sqrt{u}}\ du$
=$\int\frac{-1}{4\sqrt{u}}+\frac{\sqrt{u}}{4}\ du$
=$\frac{-\sqrt{u}}{2}+\frac{u\sqrt{u}}{6}+C$
=$\frac{-\sqrt{1-x^4}}{2}+\frac{(1-x^4)\sqrt{1-x^4}}{6}+C$
23) $\int ln(1+x)\ dx$
Jawab : misalkan $u=ln(1+x)$
                                $du=\frac{1}{1+x}\ dx$
                                $dv=dx$
                                $v=x$
$\int ln(1+x)\ dx=\int u\ dv=uv-\int v\ du$
=$ln(1+x).x-\int \frac{x}{1+x}\ dx$
=$x.ln(1+x)-\int \frac{x+1-1}{1+x}\ dx$
=$x.ln(1+x)-\int 1\ dx+\int \frac{1}{1+x}\ dx$
=$x.ln(1+x)-x+ln(1+x)+C$
24) $\int x.sec^{-1}\ x\ dx$
Jawab : misalkan $u=sec^{-1}\ x$
                                $du=\frac{1}{x\sqrt{x^2-1}}\ dx$
                                $dv=x\ dx$
                                $v=\frac{x^2}{2}$
$\int x.sec^{-1}\ x\ dx=\int u\ dv=uv-\int v\ du$
=$sec^{-1}\ x.\frac{x^2}{2}-\int\frac{x}{2\sqrt{x^2-1}}\ dx$
Obs : $\int\frac{x}{2\sqrt{x^2-1}}\ dx$
misalkan $y=x^2-1$
                  $dy=2x\ dx$ atau $\frac{dy}{2}=x\ dx$
$\int\frac{x}{2\sqrt{x^2-1}}\ dx$
=$\int\frac{1}{4\sqrt{y}}\ dy$
=$\frac{\sqrt{y}}{2}$
=$\frac{\sqrt{x^2-1}}{2}$
Jadi,
$\int x.sec^{-1}\ x\ dx$
=$sec^{-1}\ x.\frac{x^2}{2}-\int\frac{x}{2\sqrt{x^2-1}}\ dx$
=$sec^{-1}\ x.\frac{x^2}{2}-\frac{\sqrt{x^2-1}}{2}+C$
25) $\int\sqrt{x^2+9}\ dx$
Jawab : misalkan $x=3tan\ u$
                                $dx=3sec^2\ u\ du$
$\int\sqrt{x^2+9}\ dx$
=$\int\sqrt{9tan^2\ u+9}.3sec^2\ u\ du$
=$\int 9sec^3\ u\ du$
=$\frac{9}{2}\left(sec\ u.tan\ u+ln|tan\ u+sec\ u|\right)+C$
=$\frac{9}{2}\left(\frac{\sqrt{x^2+9}}{3} .\frac{x}{3}+ln|\frac{x}{3}+\frac{\sqrt{x^2+9}}{3}|\right)+C$



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