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Integral with ax+bax+b

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∫1ax+b dx=1aln⁡∣ax+b∣+C\int {\frac {1}{ax+b}}\ \mathrm{d}x=\frac {1}{a}\ln \left|{ax+b}\right|+C
Derivation
I=∫1ax+b dx=∫1ax+b d(ax+b)a(Direct Substitution)=1a∫1ax+b d(ax+b)=1aln⁡∣ax+b∣+C\begin{align*} I&= \int {\frac{1}{ax+b}}\ \mathrm{d}x\\ &=\int {\frac{1}{ax+b}}\ \frac{\mathrm{d}(ax+b)}{a} &&\text{(Direct Substitution)}\\ &=\frac{1}{a} \int {\frac{1}{ax+b}}\ \mathrm{d}(ax+b)\\ &=\frac{1}{a} \ln{\left|ax+b \right|}+C \tag*{$\blacksquare$} \end{align*}

Direct Substitution

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∫(ax+b)μ dx=1a(μ+1)(ax+b)μ+1+C (μ≠−1)\int\left(ax+b\right)^{\mu}\ \mathrm{d}x=\frac {1}{a\left(\mu+1\right)}\left(ax+b\right)^{\mu+1}+C\ \left(\mu\neq-1\right)
Derivation
I=∫(ax+b)μ dx=∫(ax+b)μ d(ax+b)a(Direct Substitution)=1a∫(ax+b)μ d(ax+b)=1a(ax+b)μ+1(μ+1)+C=1a(μ+1)(ax+b)μ+1+C\begin{align*} I&= \int{\left(ax+b\right)^{\mu}}\ \mathrm{d}x\\ &=\int{\left(ax+b\right)^{\mu}}\ \frac{\mathrm{d}(ax+b)}{a} &&\text{(Direct Substitution)}\\ &=\frac{1}{a} \int {\left(ax+b\right)^{\mu}}\ \mathrm{d}(ax+b)\\ &=\frac{1}{a} \frac{\left(ax+b\right)^{\mu+1}}{\left(\mu+1\right)}+C\\ &=\frac{1}{a(\mu+1)} \left(ax+b\right)^{\mu+1}+C \tag*{$\blacksquare$} \end{align*}

Direct Substitution

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∫xax+b dx=1a2(ax−bln⁡∣ax+b∣)+C\int\frac {x}{ax+b}\ \mathrm{d}x=\frac {1}{a^2}\left( ax-b\ln \left|{ax+b}\right|\right)+C
Derivation
I=∫xax+b dx=∫ax+b−ba(ax+b) dx=1a∫(1−bax+b) dx=1a(x−ba∣ax+b∣+C)(Integral Formula #1)=1a2(ax−bln⁡∣ax+b∣)+C\begin{align*} I&=\int{\frac{x}{ax+b}}\ \mathrm{d}x\\ &=\int{\frac{ax+b-b}{a\left(ax+b\right)}}\ \mathrm{d}x\\ &=\frac{1}{a} \int{\left(1-\frac{b}{ax+b}\right)}\ \mathrm{d}x\\ &=\frac{1}{a} \left(x- \frac{b}{a} \left|ax+b \right|+C\right) &&\text{(Integral Formula \#1)}\\ &=\frac {1}{a^2}\left( ax-b\ln \left|{ax+b}\right|\right)+C \tag*{$\blacksquare$} \end{align*}

Integral Formula #1

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∫x2ax+b dx=1a3[12(ax−b)2+b2ln⁡∣ax+b∣]+C\int\frac{x^2}{ax+b}\ \mathrm{d}x=\frac{1}{a^3}\left[\frac{1}{2}\left(ax-b\right)^2+b^2\ln\left|{ax+b}\right|\right]+C
Derivation
I=∫x2ax+b dx=∫a2x2−b2+b2a2(ax+b) dx=1a2∫((ax+b)(ax−b)ax+b+b2ax+b) dx=1a2[∫(ax−b) d(ax+b)a+∫b2ax+b dx](Direct Substitution)=1a2[(ax−b)22a+b2aln⁡∣ax+b∣+C](Integral Formula #1)=1a3[12(ax−b)2+b2ln⁡∣ax+b∣]+C\begin{align*} I&=\int{\frac{x^2}{ax+b}}\ \mathrm{d}x\\ &=\int{\frac{a^2x^2-b^2+b^2}{a^2\left(ax+b\right)}}\ \mathrm{d}x\\ &=\frac{1}{a^2} \int{\left(\frac{\cancel{\left(ax+b \right)}\left(ax-b \right)}{\cancel{ax+b}}+\frac{b^2}{ax+b}\right)}\ \mathrm{d}x\\ &=\frac{1}{a^2} \left[\int{\left(ax-b\right)}\ \frac{\mathrm{d}(ax+b)}{a}+\int{\frac{b^2}{ax+b}}\ \mathrm{d}x\right] &&\text{(Direct Substitution)}\\ &=\frac{1}{a^2} \left[\frac{\left(ax-b\right)^2}{2a}+ \frac{b^2}{a}\ln \left|{ax+b}\right|+C \right] &&\text{(Integral Formula \#1)}\\ &=\frac{1}{a^3}\left[\frac{1}{2}\left(ax-b\right)^2+b^2\ln\left|{ax+b}\right|\right]+C \tag*{$\blacksquare$} \end{align*}

Direct Substitution

Integral Formula #1

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∫1x(ax+b) dx=1bln⁡∣xax+b∣+C\int \frac {1}{x\left(ax+b\right)}\ \mathrm{d}x=\frac {1}{b}\ln {\left|\frac {x}{ax+b}\right|}+C
Derivation
I=∫1x(ax+b) dx=∫(Ax+Bax+b) dx(Partial Fraction Decomposition)=∫A(ax+b)+Bxx(ax+b) dx=∫(Aa+B)x+Abx(ax+b) dx\begin{align*} I&=\int{\frac {1}{x\left(ax+b\right)}}\ \mathrm{d}x\\ &=\int{\left(\frac{A}{x}+\frac{B}{ax+b}\right)}\ \mathrm{d}x &&\text{(Partial Fraction Decomposition)}\\ &=\int{\frac {A\left(ax+b\right)+Bx}{x\left(ax+b\right)}}\ \mathrm{d}x\\ &=\int{\frac {\left(Aa+B\right)x+Ab}{x\left(ax+b\right)}}\ \mathrm{d}x \end{align*}

Comparing coefficients gives:

(Aa+B)x+Ab≡1  ⟹  {Aa+B=0Ab=1  ⟹  {A=1bB=−ab\left(Aa+B\right)x+Ab\equiv 1 \implies \left\{ \begin{align*} &Aa+B=0\\ &Ab=1 \end{align*} \right. \implies \left\{ \begin{align*} &A=\frac{1}{b}\\ &B=-\frac{a}{b} \end{align*} \right.

Plug in back:

I=∫(1bx+−abax+b) dx=∫[1bx−ab(ax+b)]dx=1b[∫1x dx−∫a(ax+b) dx]=1b(ln⁡∣x∣−ln⁡∣ax+b∣+C)(#1)=1bln⁡∣xax+b∣+C\begin{align*} I&=\int{\left(\frac{\frac{1}{b}}{x}+\frac{-\frac{a}{b}}{ax+b}\right)}\ \mathrm{d}x\\ &=\int{\left[ \frac{1}{bx}-\frac{a}{b\left(ax+b\right)} \right]}\mathrm{d}x\\ &=\frac{1}{b}\left[ \int{\frac{1}{x}}\ \mathrm{d}x- \int{\frac{a}{\left(ax+b\right)}}\ \mathrm{d} x\right]\\ &=\frac{1}{b}\left(\ln{\left|x \right|}-\ln{\left|ax+b \right|}+C\right) &&\text{(\#1)}\\ &=\frac {1}{b}\ln {\left|\frac {x}{ax+b}\right|}+C \tag*{$\blacksquare$} \end{align*}

Integral Formula #1

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∫1x2(ax+b) dx=−1bx+ab2ln⁡∣ax+bx∣+C\int \frac{1}{x^2\left(ax+b\right)}\ \mathrm{d}x=-\frac{1}{bx}+\frac{a}{b^2}\ln \left|\frac {ax+b}{x} \right|+C
Derivation
I=∫1x2(ax+b) dx=∫(Ax+Bx2+Cax+b) dx(Partial Fraction Decomposition)=∫Ax(ax+b)+B(ax+b)+Cx2x2(ax+b) dx=∫(Aa+C)x2+(Ab+Ba)x+Bbx2(ax+b) dx\begin{align*} I&=\int{\frac {1}{x^2\left(ax+b\right)}}\ \mathrm{d}x\\ &=\int{\left(\frac{A}{x}+\frac{B}{x^2}+\frac{C}{ax+b}\right)}\ \mathrm{d}x &&\text{(Partial Fraction Decomposition)}\\ &=\int{\frac {Ax\left(ax+b\right)+B\left(ax+b\right)+Cx^2}{x^2\left(ax+b\right)}}\ \mathrm{d}x\\ &=\int{\frac {\left(Aa+C\right)x^2+\left(Ab+Ba\right)x+Bb}{x^2\left(ax+b\right)}}\ \mathrm{d}x\\ \end{align*}

Comparing coefficients gives:

(Aa+C)x2+(Ab+Ba)x+Bb≡1  ⟹  {Aa+C=0Ab+Ba=0Bb=1  ⟹  {A=−ab2B=1bC=a2b2\left(Aa+C\right)x^2+\left(Ab+Ba\right)x+Bb\equiv 1 \implies \left\{ \begin{align*} &Aa+C=0\\ &Ab+Ba=0\\ &Bb=1 \end{align*} \right. \implies \left\{ \begin{align*} &A=-\frac{a}{b^2}\\ &B=\frac{1}{b}\\ &C=\frac{a^2}{b^2} \end{align*} \right.

Plug in back:

I=∫(−ab2x+1bx2+a2b2ax+b) dx=∫[−ab2x+1bx2+a2b2(ax+b)]dx=1b∫1x2 dx+ab2[−∫1x dx+∫a(ax+b) dx]=−1bx+ab2(−ln⁡∣x∣+ln⁡∣ax+b∣+C)(Integral Formula #1)=−1bx+ab2ln⁡∣ax+bx∣+C\begin{align*} I&=\int{\left(\frac{-\frac{a}{b^2}}{x}+\frac{\frac{1}{b}}{x^2}+\frac{\frac{a^2}{b^2}}{ax+b}\right)}\ \mathrm{d}x\\ &=\int{\left[ -\frac{a}{b^2x}+\frac{1}{bx^2}+\frac{a^2}{b^2\left(ax+b\right)} \right]}\mathrm{d}x\\ &=\frac{1}{b}\int{\frac{1}{x^2}}\ \mathrm{d}x+ \frac{a}{b^2}\left[ -\int{\frac{1}{x}}\ \mathrm{d}x+ \int{\frac{a}{\left(ax+b\right)}}\ \mathrm{d} x\right]\\ &=-\frac{1}{bx}+\frac{a}{b^2}\left(-\ln{\left|x \right|}+\ln{\left|ax+b \right|}+C\right) &&\text{(Integral Formula \#1)}\\ &=-\frac{1}{bx}+\frac {a}{b^2}\ln \left|\frac {ax+b}{x}\right|+C \tag*{$\blacksquare$} \end{align*}

Integral Formula #1

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∫x(ax+b)2  dx=1a2(ln⁡∣ax+b∣+bax+b)+C\int \frac{x}{\left(ax+b\right)^2} \, \ \mathrm{d}x = \frac{1}{a^2} \left( \ln\left|ax+b\right| + \frac{b}{ax+b} \right) + C
Derivation
I=∫x(ax+b)2 dx=∫ax+b−ba(ax+b)2 dx=1a∫[ax+b(ax+b)2−b(ax+b)2] dx=1a[∫1(ax+b) dx−∫b(ax+b)2 d(ax+b)a](Direct Substitution)=1a[1aln⁡∣ax+b∣+ba(ax+b)+C](Integral Formula #1)=1a2(ln⁡∣ax+b∣+bax+b)+C\begin{align*} I&=\int{\frac{x}{\left(ax+b\right)^2}}\ \mathrm{d}x\\ &=\int{\frac{ax+b-b}{a\left(ax+b\right)^2}} \ \mathrm{d}x\\ &=\frac{1}{a} \int{\left[\frac{ax+b}{\left(ax+b \right)^2}-\frac{b}{\left(ax+b\right)^2}\right]}\ \mathrm{d}x\\ &=\frac{1}{a} \left[\int{\frac{1}{\left(ax+b\right)}}\ \mathrm{d}x-\int{\frac{b}{\left(ax+b\right)^2}}\ \frac{\mathrm{d}\left(ax+b\right)}{a}\right] &&\text{(Direct Substitution)}\\ &=\frac{1}{a} \left[\frac{1}{a}\ln\left|ax+b\right|+ \frac{b}{a\left(ax+b\right)}+C \right] &&\text{(Integral Formula \#1)}\\ &=\frac{1}{a^2} \left( \ln\left|ax+b\right| + \frac{b}{ax+b} \right) + C \tag*{$\blacksquare$} \end{align*}

Direct Substitution

Integral Formula #1

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∫x2(ax+b)2 dx=1a3(ax−2bln⁡∣ax+b∣−b2ax+b)+C\int \frac{x^2}{\left(ax+b\right)^2}\ \mathrm{d}x=\frac {1}{a^3} \left(ax-2b\ln{\left|ax+b\right|}-\frac{b^2}{ax+b} \right )+C
Derivation
I=∫x2(ax+b)2 dx=∫a2x2−b2+b2a2(ax+b)2 dx=1a2∫[(ax+b)(ax−b)(ax+b)2+b2(ax+b)2] dx=1a2[∫ax+b−2bax+b dx+∫b2(ax+b)2 dx]=1a2[∫(1−2bax+b) dx+∫b2(ax+b)2 d(ax+b)a](Direct Substitution)=1a2[x−2baln⁡∣ax+b∣+b2a(ax+b)+C](Integral Formula #1)=1a3(ax−2bln⁡∣ax+b∣−b2ax+b)+C\begin{align*} I&=\int{\frac{x^2}{\left(ax+b\right)^2}}\ \mathrm{d}x\\ &=\int{\frac{a^2x^2-b^2+b^2}{a^2\left(ax+b\right)^2}} \ \mathrm{d}x\\ &=\frac{1}{a^2} \int{\left[\frac{\left(ax+b\right)\left(ax-b\right)}{\left(ax+b \right)^2}+\frac{b^2}{\left(ax+b\right)^2}\right]}\ \mathrm{d}x\\ &=\frac{1}{a^2} \left[\int{\frac{ax+b-2b}{ax+b}}\ \mathrm{d}x+\int{\frac{b^2}{\left(ax+b\right)^2}}\ \mathrm{d}x\right]\\ &=\frac{1}{a^2} \left[\int{\left(1-\frac{2b}{ax+b}\right)}\ \mathrm{d}x+\int{\frac{b^2}{\left(ax+b\right)^2}}\ \frac{\mathrm{d}\left(ax+b\right)}{a}\right] &&\text{(Direct Substitution)}\\ &=\frac{1}{a^2} \left[x-\frac{2b}{a}\ln\left|ax+b\right|+ \frac{b^2}{a\left(ax+b\right)}+C \right] &&\text{(Integral Formula \#1)}\\ &=\frac {1}{a^3} \left(ax-2b\ln{\left|ax+b\right|}-\frac{b^2}{ax+b} \right)+C \tag*{$\blacksquare$} \end{align*}

Direct Substitution

Integral Formula #1

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∫1x(ax+b)2 dx=1b(ax+b)+1b2ln⁡∣xax+b∣+C\int \frac{1}{x\left(ax+b\right)^2}\ \mathrm{d}x=\frac{1}{b\left(ax+b\right)}+\frac{1}{b^2}\ln {\left|\frac{x}{ax+b} \right|}+C
Derivation
I=∫1x(ax+b)2 dx=∫[Ax+B(ax+b)+C(ax+b)2] dx(Partial Fraction Decomposition)=∫A(ax+b)2+Bx(ax+b)+Cxx(ax+b)2 dx=∫Aa2x2+2Aabx+Ab2+Bax2+Cxx(ax+b)2 dx=∫(Aa2+Ba)x2+(2Aab+C)x+Ab2x(ax+b)2 dx\begin{align*} I&=\int{\frac {1}{x\left(ax+b\right)^2}}\ \mathrm{d}x\\ &=\int{\left[\frac{A}{x}+\frac{B}{\left(ax+b\right)}+\frac{C}{\left(ax+b\right)^2}\right]}\ \mathrm{d}x &&\text{(Partial Fraction Decomposition)}\\ &=\int{\frac {A\left(ax+b\right)^2+Bx\left(ax+b\right)+Cx}{x\left(ax+b\right)^2}}\ \mathrm{d}x\\ &=\int{\frac {Aa^2x^2+2Aabx+Ab^2+Bax^2+Cx}{x\left(ax+b\right)^2}}\ \mathrm{d}x\\ &=\int{\frac {\left(Aa^2+Ba\right)x^2+\left(2Aab+C\right)x+Ab^2}{x\left(ax+b\right)^2}}\ \mathrm{d}x\\ \end{align*}

Comparing coefficients gives:

(Aa2+Ba)x2+(2Aab+C)x+Ab2≡1  ⟹  {Aa2+Ba=02Aab+C=0Ab2=1  ⟹  {A=1b2B=−ab2C=−ab\left(Aa^2+Ba\right)x^2+\left(2Aab+C\right)x+Ab^2\equiv 1 \implies \left\{ \begin{align*} &Aa^2+Ba=0\\ &2Aab+C=0\\ &Ab^2=1 \end{align*} \right. \implies \left\{ \begin{align*} &A=\frac{1}{b^2}\\ &B=-\frac{a}{b^2}\\ &C=-\frac{a}{b} \end{align*} \right.

Plug in back:

I=∫[1b2x+−ab2(ax+b)+−ab(ax+b)2] dx=∫[1b2x−ab2(ax+b)−ab(ax+b)2]dx=ab∫−1(ax+b)2 d(ax+b)a+1b2[∫1x dx−∫a(ax+b) dx](Direct Substitution)=1b(ax+b)+1b2(ln⁡∣x∣−ln⁡∣ax+b∣+C)(Integral Formula #1)=1b(ax+b)+1b2ln⁡∣xax+b∣+C\begin{align*} I&=\int{\left[\frac{\frac{1}{b^2}}{x}+\frac{-\frac{a}{b^2}}{\left(ax+b\right)}+\frac{-\frac{a}{b}}{\left(ax+b\right)^2}\right]}\ \mathrm{d}x\\ &=\int{\left[\frac{1}{b^2x}-\frac{a}{b^2\left(ax+b\right)}-\frac{a}{b\left(ax+b\right)^2}\right]}\mathrm{d}x\\ &=\frac{\cancel{a}}{b}\int{-\frac{1}{\left(ax+b\right)^2}}\ \frac{\mathrm{d}\left(ax+b\right)}{\cancel{a}}+ \frac{1}{b^2}\left[\int{\frac{1}{x}}\ \mathrm{d}x- \int{\frac{a}{\left(ax+b\right)}}\ \mathrm{d} x\right]&&\text{(Direct Substitution)}\\ &=\frac{1}{b\left(ax+b\right)}+\frac{1}{b^2}\left(\ln{\left|x \right|}-\ln{\left|ax+b \right|}+C\right) &&\text{(Integral Formula \#1)}\\ &=\frac{1}{b\left(ax+b\right)}+\frac{1}{b^2}\ln {\left|\frac{x}{ax+b} \right|}+C \tag*{$\blacksquare$} \end{align*}

Direct Substitution

Integral Formula #1