Evaluate $\displaystyle\int \dfrac{5x}{(x+1)(x^2+9)}\,dx$.

OR

Evaluate $\displaystyle\int_0^\pi \dfrac{x\sin x}{1+\cos^2 x}\,dx$.

Question

Evaluate $\displaystyle\int \dfrac{5x}{(x+1)(x^2+9)}\,dx$.

OR

Evaluate $\displaystyle\int_0^\pi \dfrac{x\sin x}{1+\cos^2 x}\,dx$.

Accepted Answer

> [!TLDR]
> Resolve into partial fractions with an unknown linear factor over $x^2+9$, then integrate each piece.
>
> $$\dfrac{5x}{(x+1)(x^2+9)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+9}$$
>
> $5x = A(x^2+9)+(Bx+C)(x+1)$.
>
> Put $x=-1$: $-5=10A \implies A=-\dfrac12$.
>
> Comparing $x^2$ coefficients: $0=A+B \implies B=\dfrac12$.
>
> Comparing constants: $0=9A+C \implies C=\dfrac92$.
>
> (Check $x^1$: $5=B+C=	frac12+	frac92=5$ ✓)
>
> So:
> $$\int\dfrac{5x}{(x+1)(x^2+9)}dx = -\dfrac12\int\dfrac{dx}{x+1}+\dfrac12\int\dfrac{x\,dx}{x^2+9}+\dfrac92\int\dfrac{dx}{x^2+9}$$
>
> $$=-\dfrac12\ln|x+1|+\dfrac14\ln(x^2+9)+\dfrac92\cdot\dfrac13	an^{-1}\dfrac{x}{3}+c$$
>
> $$=-\dfrac12\ln|x+1|+\dfrac14\ln(x^2+9)+\dfrac32	an^{-1}\dfrac{x}{3}+c$$
>
> ---
> **OR: Evaluate $\displaystyle\int_0^\pi \dfrac{x\sin x}{1+\cos^2 x}dx$**
>
> Let $I=\displaystyle\int_0^\pi \dfrac{x\sin x}{1+\cos^2x}dx$. Using $\int_0^a f(x)dx=\int_0^a f(a-x)dx$:
> $$I=\int_0^\pi\dfrac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\dfrac{\sin x}{1+\cos^2x}dx - I$$ …

Q.Evaluate $\displaystyle\int \dfrac{5x}{(x+1)(x^2+9)}\,dx$ .

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Q.Evaluate ∫5x(x+1)(x2+9) dx\displaystyle\int \dfrac{5x}{(x+1)(x^2+9)}\,dx∫(x+1)(x2+9)5x​dx.

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Q.Evaluate $\displaystyle\int \dfrac{5x}{(x+1)(x^2+9)}\,dx$ .

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